Đề thi thử THPT quốc gia

87 0 0
  • Loading ...
    Loading ...
    Loading ...

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

Tài liệu liên quan

Thông tin tài liệu

Ngày đăng: 24/02/2021, 06:17

Định nghĩa 1 (Giới hạn của hàm số tại một điểm)... Phương pháp giải:.[r] (1) CHƯƠNG 4 GIỚI HẠN BÀI GIỚI HẠN CỦA HÀM SỐ A TÓM TẮT LÝ THUYẾT Định nghĩa 1 (Giới hạn hàm số điểm) Giả sử (a b; ) khoảng chứa điểm x0 f hàm số xác định tập hợp (a b; )  \ x0 Ta nói hàm số f có giới hạn số thực L x dần đến x0 (hoặc điểm x0) với dãy số ( )xn tập hợp (a b; )  \ x0 mà limxn =x0 ta có lim f x( )n =L Khi ta viết ( ) 0 lim xx f x =L f x( ) Lxx0 Định nghĩa 2 (Giới hạn hàm số vô cực) Giả sử hàm số f xác định khoảng (a;+) Ta nói hàm số f có giới hạn số thực L x dần tới + với dãy số ( )xn khoảng (a;+) mà limxn = + ta có ( ) lim f xn =L Khi ta viết lim ( ) x→+f x =L f x( )→L x→ + GIỚI HẠN HỮA HẠN GIỚI HẠN VÔ CỰC Giới hạn đặc biệt 1) 0 0 lim xx x=x 2) 0 lim xx c=c (c ) Giới hạn đặc biệt 1) lim k x→+x = + 2) limx k c x → = 3) 0 1 lim x→− x = − 4) 0 1 lim x→+ x = + 5) li ( 0) m x k k k k x →− +  =  −       Định lí Nếu ( ) 0 lim xx f x =L ( ) lim xx g x =M 1) ( ) ( ) 0 lim xx f xg x = L M 2) ( ) ( ) 0 lim xx f x g x =L M 3) ( ) ( ) 0 lim x x f x L g x M → = với M 0 Nếu f x( )0 ( ) 0 lim xx f x =L ( ) 0 lim xx f x = L xlim→x0 f x( )= L Định lí Nếu ( ) 0 lim xx f x = L ( ) lim xx f x =  ( ) ( ) ( )( ) 0 lim lim lim x x x x x x L g x f x g x L g x → → → +    =      −   Nếu ( ) 0 lim xx g x = ( ) ( ) ( ) ( ) 0 lim x x L g x f x g x L g x → +   =  −  (2)Page 2 Giới hạn bên ( ) ( ) ( ) 0 0 lim l i i l m m x x x x x x f x f x L f x L + − → =  → = → = B DẠNG TOÁN VÀ BÀI TẬP Dạng Tính giới hạn vô định dạng 0 0, đó tử thức và mẫu thức là các đa thức Phương pháp giải: Khử dạng vô định cách phân tích thành tích cách chia Hooc – nơ (đầu rơi, nhân tới, cộng chéo), rời sau đơn giản biểu thức để khử dạng vơ định  VÍ DỤ Ví dụ 1. Tính giới hạn 2 2 2 14 lim 4 x x x A x → + − = − Đs: 11 A= Lời giải Ta có 2 2 2 7 2(x 2)(x ) 2 14 2 11 lim lim lim 4 (x 2)(x 2) x x x x x x A x x → → → − + + − + = = = = − − + + ! Cần nhớ: ( )( ) 1 ( ) a f x = x +bx+ =c a xx xx với x x1, 2là nghiệm phương trình ( ) f x = Học sinh thường quên nhân thêm a Ví dụ Tính giới hạn 3 3 2 2 lim 4 13 x x x x A x x x → − − − = − + − Đs: 11 17 A= Lời giải ( )( ) ( )( ) 2 3 2 3 2 3 3 3 2 11 lim lim lim 4 13 3 4 17 x x x x x x x x x x x A x x x x x x x x → → → − + + − − − + + = = = = − + − − − + − + Nhận xét:Bảng chia Hooc – nơ (đầu rơi, nhân tới cộng chéo) sau: Phân tích 2x3−5x2−2x−3thành tích số: ( )( ) 3 2 2x 5x 2x x 2x x  − − − = − + + Phân tích (3)( )( ) 3 2 4x 13x 4x x 4x x  − + − = − − + Ví dụ Tính giới hạn 100 50 2 lim 2 x x x A x x → − + = − + Đs: 49 24 A= Lời giải Ta có ( ) ( ) ( ) ( ) 99 100 100 50 50 49 1 1 1 2 ( ) ( 1) lim lim lim 2 ( ) ( 1) 1 x x x x x x x x x x x A x x x x x x x x → → → − − − − + − − − = = = − + − − − − − − ( )( ) ( ) ( )( ) ( ) ( )( ) ( )( ) 98 97 96 48 47 46 1 99 98 97 49 48 47 1 1 1 lim 1 1 1 lim 1 x x x x x x x x x x x x x x x x x x x x x x x x x x x x → → − + + + + + − − = − + + + + + − − − + + + + + − = − + + + + + − ( ) ( ) 99 98 97 49 48 47 1 98 49 lim 48 24 x x x x x x x x x x x → + + + + + − = = = + + + + + − !Cần nhớ:Hằng đẳng thức ( )( 2 ) 1 1 n n n x − = xx − +x − + +x + +x Chứng minh: Xét cấp số nhân 1, ,x x x2, 3, ,xn−1có n số hạng và u1=1,q=x Khi đó ( )( ) 2 1 1 1 1 1 1 n n n n n n q x S x x x u x x x x x q x − − − − = + + + + = =  − = − + + + + − −  BÀI TẬP ÁP DỤNG Bài 1. Tính giới hạn sau: 1) 2 2 3 lim 4 x x x A x → − + = − ĐS: 1 A= 2) 2 1 lim 3 x x A x x → − = + − ĐS: 2 A= 3) 2 7 12 lim 9 x x x A x → − + = − ĐS: 1 A= − 4) 2 9 20 lim 5 x x x A x x → − + = − ĐS: 1 A= 5) 2 3 10 lim 5 x x x A x x → − + = − + ĐS: A=8 6) 2 2 lim 2 x x x A x x → + − = − − ĐS: 4 A= 7) 4 2 16 lim 6 x x A x x →− − = + + ĐS: A= −16 8) 2 lim 5 x x x A x x → − − = − + ĐS: 4 A= − 9) 3 2 8 lim 3 x x A x x → − = − + ĐS: A=12 10) 3 2 8 lim 11 18 x x A x x →− + = + + ĐS: 12 (4)Page 4 1) 3 2 2 lim 1 x x x x A x → − + + = − ĐS: A= −1 2) 3 3 lim 4 x x x A x x → − + = − + ĐS: 1 A= 3) 3 3 1 2 lim 1 x x x x A x x x →− + + + = + − − ĐS: 1 A= 4) 4 3 1 1 lim 5 x x x x A x x x → − − + = − + − ĐS: 3 A= − 5) 3 2 2 lim 3 x x x x A x →− − + + + = − ĐS: 18 19 A= + 6) 3 4 3 5 lim 8 x x x x A x x → − + + = − − ĐS: A=0 7) 3 4 1 1 lim 4 x x A x x → − = − + ĐS: 3 A= 8) 3 2 1 12 lim 2 x A x x →   =  −  − −   ĐS: 1 A= 9) 2 2 2 1 lim 3 x A x x x x →   =  +  − − − −   ĐS: A= −2 10) 2 3 1 1 lim 2 x A x x x →   =  −  + − −   ĐS: 1 A= Bài 3. Tính giới hạn sau: 1) 20 30 2 lim 2 x x x A x x → − + = − + ĐS: 8 14 A= 2) 50 1 lim 3 x x A x x → − = − + ĐS: A= −50 3) ( )2 1 lim 1 n x x nx n A x → − + − = − (Với n là số nguyên) ĐS: 2 n n A= − 4) ( ) ( ) 1 2 1 lim 1 n x x n x n A x + → − + + = − ĐS: ( 1) 2 n n A= + 5) 2 2 1 lim n m x x x x x n A x x x x m → + + + + − = + + + + − (m n, là số nguyên) ĐS: ( ) ( ) 1 n n A m m + = + 6) 1 lim 1 m n x m n A x x →   =  −  − −   ĐS: m n A= −  LỜI GIẢI Bài 1. 1) Ta có ( )( ) ( )( ) 2 2 2 1 3 1 lim lim lim 4 2 x x x x x x x x A x x x x → → → − − − + − = = = = − − + + 2) Ta có ( )( ) ( )( ) 2 1 1 1 1 lim lim lim 3 4 x x x x x x x A x x x x x → → → − + − + = = = = + − − + + 3) Ta có ( )( ) ( )( ) 2 3 3 3 7 12 lim lim lim 9 3 x x x x x x x x A x x x x → → → − − − + − = = = = − (5)4) Ta có ( )( ) ( ) 2 5 5 4 9 20 lim lim lim 5 5 x x x x x x x x A x x x x x → → → − − − + − = = = = − − 5) Ta có ( )( ) ( )( ) 2 3 3 3 3 10 3 lim lim lim 5 x x x x x x x x A x x x x x → → → − − − + − = = = = − + − − − 6) Ta có ( )( ) ( )( ) 2 1 1 1 2 3 lim lim lim 2 1 2 x x x x x x x x A x x x x x → → → − + + − + = = = = − − − + + 7) Ta có ( )( )( ) ( )( ) ( )( ) ( ) 2 4 2 2 2 4 16 lim lim lim 16 6 4 x x x x x x x x x A x x x x x →− →− →− − + + − + − = = = = − + + + + + 8) Ta có ( )( ) ( )( ) (( )) 1 1 1 3 2 lim lim lim 3 5 4 x x x x x x x x A x x x x x → → → − + + − − = = = = − − + − − − 9) Ta có ( )( ) ( )( ) ( ( ) ) 2 3 2 2 2 4 8 lim lim lim 12 3 2 1 x x x x x x x x x A x x x x x → → → − + + + + − = = = = − + − − − ! Cần nhớ: Hằng đẳng thức a3+b3 =(a+b)(a2−ab b+ 2)và a3−b3=(a b− )(a2+ab b+ 2) 10) Ta có ( )( ) ( )( ) ( ( ) ) 2 3 2 2 2 4 8 12 lim lim lim 11 18 9 x x x x x x x x x A x x x x x →− →− →− + − + − + + = = = = + + + + + Bài 2. 1) ( )( ) ( )( ) 2 3 2 2 1 1 1 2 2 lim lim lim 1 1 x x x x x x x x x x x A x x x x → → → − − − − + + − − = = = = − − − + + 2) ( ) ( ) ( ) ( ) 2 2 4 2 1 1 1 3 2 lim lim lim 4 3 x x x x x x x x A x x x x x x x → → → − + − + + = = = = − + − + + + + 3) ( ) ( ) ( ) ( ) 2 3 2 3 1 1 1 2 1 lim lim lim 1 1 1 x x x x x x x x x A x x x x x x →− →− →− + + + + + + = = = = + − − + − − 4) ( ) ( ) ( ) ( ) 2 2 4 2 3 1 1 1 1 lim lim lim 5 3 x x x x x x x x x x x A x x x x x x → → → − + + − − + + + = = = = − − + − − − − 5) Ta có ( )( ( ) ) ( )( ) 2 3 2 3 3 3 3 2 lim lim 3 3 3 x x x x x x x x A x x x →− →−  + − + + +  − + + +   = = −  −  + −    ( ) 2 3 2 3 3 18 19 3 lim 6 x x x x →−  − + + +  +   = − =  −    6) Ta có ( )( ) ( ) ( )(( )) 2 3 1 3 5 limx x x lim x x lim x x (6)Page 6 7) Ta có ( )( ) ( )( ) ( ( ) ) 2 3 4 3 1 1 1 1 1 lim lim lim 4 3 3 x x x x x x x x x A x x x x x x x x x → → → − − − − − − − − = = = = − + − + − − + − − 8) Ta có ( )( ) 3 3 2 1 12 12 16 lim lim 2 8 x x x x A x x x x → → − +   =  − = − − − −   ( )( ) ( ) ( ) 2 2 2 2 4 lim lim 2 2 x x x x x x x x x x → → + − + = = = + + − + + 9) Ta có ( )( ) 2 2 2 2 1 lim lim 3 6 x x x x x x A x x x x x x x x → → − − + − −   =  + = − − − − − − − −   ( ) ( ) ( )( ) ( )( ) 2 2 2 2 2 lim lim 3 2 x x x x x x x x → → − = = = − − − − − − 10) Ta có ( )( ) ( )( ) 3 2 3 1 1 1 1 lim lim lim 2 2 x x x x x x x x x A x x x x x x x x x → → → − − − + − − +   =  − = = + − − + − − + − −   ( ) ( ) ( ) ( )( ) ( )( ) 2 2 2 1 1 1 lim lim 9 2 1 x x x x x x x x x x x x → → − + + = = = + + + − + + + Bài 3. 1) Ta có ( ) ( ) ( ) ( ) ( ) ( ) 19 20 20 30 30 29 1 1 1 1 lim lim lim 2 1 1 x x x x x x x x x x x A x x x x x x x x → → → − − − − − − − + = = = − + − − − − − − ( )( ) ( ) ( )( ) ( ) ( )( ) ( )( ) 18 17 19 18 28 27 29 28 1 1 1 lim lim 1 1 x x x x x x x x x x x x x x x x x x x x x x → → − + + + + − − − + + + − = = − + + + + − − − + + + − ( ) ( ) 19 18 29 28 1 18 9 lim 28 24 x x x x x x x → + + + − = = = + + + − 2) Ta có ( )( ) ( )( ) 49 48 50 49 48 2 1 1 1 x 1 x lim lim lim 50 3 2 x x x x x x x x x A x x x x x → → → − + + + + − + + + + = = = = − − + − − − 3) Ta có ( ) ( ) ( ) ( ) 2 1 1 1 lim lim 1 n n x x x n x x nx n A x x → → − − − − + − = = − − ( )( ) ( ) ( ) 1 2 1 x 1 lim 1 n n x x x x n x x − − → − + + + + − − = − ( )( ) ( ) 1 1 2 2 1 1 x x 1 lim lim 1 n n n n x x x x x n x x n x x − − − − → → − + + + + − + + + + − = = − − 1 2 1 1 x 1 lim 1 n n x x x x x − − → − + − + + − + − = (7)( )( ) ( )( ) ( ) 1 x 1 x lim 1 n n n n x x x x x x x x x − − − − → − + + + + + − + + + + + + − = − ( ) ( ) 1 lim n n x n n x x x x x x − − − − →   =  + + + + + + + + + + +  ( 1) ( 2) 2 n n n n − = − + − + + = 4) Ta có ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 2 1 1 1 1 1 lim lim lim 1 1 n n n x x x x x n x x x n x x n x n A x x x + + → → → − − − − − − − + + = = = − − − ( )( ) ( ) ( ) ( )( ) ( ) 1 2 1 1 x 1 x lim lim 1 n n n n x x x x x x n x x x x n x x − − − → → − + + + + − − − + + + − = = − − 1 2 1 x 1 x 1 lim lim 1 n n n n x x x x x n x x x x x − − → → + + + + − − + − + + − + − = = − − ( )( ) ( )( ) ( ) 1 1 x 1 x lim 1 n n n n x x x x x x x x x − − − − → − + + + + + − + + + + + + − = − ( ) ( ) 1 lim n n x n n x x x x x x − − − − →   =  + + + + + + + + + + +  ( 1) ( 2) ( 1) 2 n n n n n + = + − + − + + = 5) Ta có 2 2 1 1 1 lim lim 1 1 n n n m m m x x x x x x n x x x x A x x x x m x x x x − − → → + + + + − − + − + + − + − = = + + + + − − + − + + − + − ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) 1 2 1 2 1 1 x 1 x lim 1 x 1 x n n n n m m m m x x x x x x x x x x x x x x x − − − − − − − − → − + + + + + − + + + + + + − = − + + + + + − + + + + + + − ( ) ( ) ( ) ( ) 1 2 1 2 1 x x lim x x n n n n m m m m x x x x x x x x x − − − − − − − − → + + + + + + + + + + + = + + + + + + + + + + + ( ) ( ) ( ) ( ) (( )) 1 1 1 lim 1 1 x n n n n n m m m m m → + − + − + + + = = + − + − + + + 6) Ta có 1 1 lim lim 1 m n m 1 n x x m n m n A x x x x x x → →         =  − =  −  − −  − − − − − −       1 1 lim lim 1 m 1 n x x m n x x x x → →     =  − −  −  − − − −     Và ( ) ( ) ( ) ( ) 2 1 1 1 x 1 x 1 lim lim lim 1 1 x m m m m m x x x m x x x x m x x x − − → → → − + + + + − + − + + −  − = =  − −  − −   ( ) ( ) ( ) ( )( ) 2 2 1 1 1 lim 1 m m x x x x x x x x x x − − →   −  + + + + + + + +  = − + + + + ( ) ( 2) 2 1 1 1 1 1 1 lim 1 m m x x x x x m m x x x m − − → + + + + + + + + + + + + − − = = = + + + + Tương tự ta có 1 1 lim 1 n x n n x x → −  − =  − −  (8)Page 8 Dạng Tính giới hạn vô định dạng 0 0, đó tử thức mẫu thức có chứa thức Phương pháp giải: Nhân lượng liên hợp để khử dạng vơ định. VÍ DỤ Ví dụ 1. Tính giới hạn 6 3 lim 6 x x B x → − + = − Đs: 1 B= − Lời giải Ta có: ( )( ) ( )( ) 6 3 3 3 lim lim 6 6 3 3 x x x x x B x x x → → − + + + − + = = − − + + ( ) ( )( ) ( )( ) 6 6 9 1 lim lim lim 6 3 3 6 3 3 x x x x x x x x x x → → → − + − − − = = = = = − + + + + − + + − + + Ví dụ 2. Tính giới hạn 3 2 3 lim 2 x x x E x → + − − = − Đs:E= −1 Lời giải Ta có 3 3 2 2 3 2 3 2 2 2 5 6 lim lim lim 2 2 x x x A B x x x x E x x x 3 2 3 3 3 3 2 lim lim 2 2 3 2 2 3 2 4 x x x x A x x x x 2 2 3 3 3 lim lim 4 3 2 2 2 x x x x x x x x 2 2 4 2 lim lim lim 2 2 2 5 6 2 2 5 6 x x x x x x B x x x x x 2 5 lim 4 2 x x x Suy 4 E A B Ví dụ Tính giới hạn 3 1 5 lim 1 x x L x →− − + = + Đs: 5 12 L= (9)Ta có: 3 1 3 3 5 5 lim lim 1 1 5 3 2 5 3 4 x x x x L x x x x 2 1 3 13 5 5 lim lim 12 5 1 5 x x x x x x x x Ví dụ Tính giới hạn 3 2 3 lim 2 x x x E x → + − − = − Đs: 1 E= − Lời giải Ta có 3 3 2 2 3 2 2 3 2 2 3 2 2 lim lim lim 2 2 x x x x x x x E x x x 2 3 3 3 lim lim 2 2 2 2 x x x x x x x x x 2 3 3 3 lim lim 2 2 2 2 x x x x x x x x x 2 2 3 3 3 3 lim lim 4 3 2 3 2 x x x x x Ví dụ Tính giới hạn 3 0 1 lim x x x F x → + + − = Đs: F 3 = Lời giải 3 0 1 1 1 lim lim x x x x x x x F x x 3 0 1 1 2 1 lim lim x x x x x x x 0 3 3 1 1 lim lim 1 1 4 x x x x x x x x x x 2 0 3 3 4 2 lim lim 3 1 1 4 x x x x x x BÀI TẬP ÁP DỤNG (10)Page 10 1) 8 8 lim 3 x x B x Đs:B 2) 2 1 4 lim 1 x x x B x Đs: 1 B 3) 2 3 2 lim 2 x x x x B x Đs: 1 B 4) 2 2 2 lim 4 x x B x Đs: 1 16 B 5) 2 2 2 lim 4 x x B x Đs: 3 16 B 6) 2 9 3 lim 9 x x B x x Đs: 1 54 B 7) 2 2 2 lim 2 10 x x B x x Đs: 1 36 B 8) 2 1 7 2 lim 1 x x x B x Đs: 1 B 9) 2 2 lim 3 x x x x B x x Đs: 5 B Bài Tính giới hạn sau: 1) 1 3 lim 8 x x x B x Đs:B 2) 3 lim 4 x x B x x Đs: 3 B 3) 2 2 lim 1 x x x B x x Đs: 1 B 4) 3 1 lim 2 x x x B x x Đs:B 5) 2 2 lim x x x x B x x Đs:B 6) 4 1 4 lim 1 x x B x Đs:B 7) 2 2 2 lim 1 x x x B x x Đs: 3 B Bài Tính giới hạn sau: 1) 0 9 16 lim x x x L x Đs: 7 24 B 2) 1 2 5 lim 1 x x x L x Đs: 4 B 3) 3 2 2 lim 3 x x x L x Đs: 5 L 4) 2 2 2 lim 2 x x x x L x Đs:L 5) 6 5 84 lim 6 x x x x L x Đs: 74 (11)6) 0 1 lim x x x L x Đs:L 7) 2 0 4 3 lim 2 x x x x L x x Đs: 5 L 8) 2 1 3 2 lim 2 x x x x L x x Đs: 17 16 L 9) 2 0 4 lim x x x L x Đs: 5 12 L 10) 2 6 lim 1 x x x x L x Đs: 11 6 L Bài Tính giới hạn sau: 1) 3 2 4 lim 2 x x L x Đs: 1 L 2) 3 0 1 lim x x L x Đs: 1 L 3) 3 3 1 lim 3 x x L x Đs: 1 L 4) 3 1 7 lim 1 x x L x Đs: 1 L 5) 3 8 2 lim 2 x x L x Đs: 5 12 L 6) 3 1 lim 2 x x L x Đs:L 7) 3 2 10 lim 3 x x x L x x Đs: 3 L 8) 3 2 8 11 lim 3 x x x L x x Đs: 7 54 L 9) 3 3 1 7 lim 1 x x x L x Đs: 1 L 10) 3 0 2 lim x x x L x Đs: 11 12 L 11) 2 2 2 11 lim 4 x x x x L x Đs: 5 72 L 12) 3 4 lim x x x L x x Đs: L Bài Tính giới hạn sau: 1) 0 1 lim n x ax F x Đs: (12)Page 12 2) 0 1 lim n m x ax bx F x Đs: a b n m 3) 0 1 lim ( 0) 1 n m x ax F ab bx Đs: am bn 4) 0 1 lim 1 n m x ax bx F x Đs: a b n m LỜI GIẢI Bài 1. 1) 8 8 8 8 lim lim lim 9 3 3 x x x x x x x x B x x x x 8 8 lim lim 8 x x x x x x 2) 2 2 1 4 4 lim lim 1 1 4 2 x x x x x x x x B x x x x 2 2 1 2 1 4 lim lim lim 4 4 1 4 x x x x x x x x x x x x x x x x 3) 2 2 3 2 3 2 lim lim 2 2 6 2 3 x x x x x x x x x x x B x x x x x 3 3 lim lim 4 2 3 2 x x x x x x x x x x x x 4) 2 2 2 2 2 2 lim lim 4 4 2 2 x x x x x B x x x 2 2 lim 2 2 x x x x x 1 lim 16 2 2 x x x 5) 2 2 2 2 2 2 4 3 2 2 lim lim lim 4 4 x x x x x x x B x x x x x 2 3 3 lim lim 16 2 2 2 x x x x x x x x 6) 2 2 9 9 3 3 lim lim lim 9 9 x x x x x x x B x x x x x x x x 1 lim 54 (13)7) 2 2 2 2 lim lim 2 10 2 2 x x x x B x x x x x 1 lim 36 2 2 x x x 8) 2 2 2 1 7 2 7 2 lim lim 1 2 x x x x x x B x x x x 2 2 2 lim 1 2 x x x x x x 1 1 lim 1 2 x x x x x x x 3 lim 3 1 2 x x x x x 9) 2 2 2 1 2 2 2 lim lim 3 3 2 2 5 2 8 x x x x x x x x B x x x x x x x 2 1 2 1 17 2 19 17 lim lim 3 2 2 x x x x x x x x x x x x x x x x 1 2 17 lim 2 2 x x x x x x Bài 2. 1) 1 1 2 8 3 lim lim lim 8 3 3 x x x x x x x x B x x x x x x 2) 1 3 lim 4 x x B x x 1 lim 1 x x x x x x 1 4 lim 3 x x x x 3 3) 2 2 lim 1 x x x B x x 2 lim 2 2 x x x x x x x 1 lim 2 2 x x x x x 1 4) 3 1 lim 2 x x x B x x 2 3 lim 3 x x x x x x x 3 2 lim 1 x x x x x 5) 2 2 lim x x x x B x x 2 1 2 2 lim 1 2 x x x x x x x x x x x 2 1 2 1 lim 1 2 x x x x x x x x x 2 1 lim 2 x x (14)Page 14 6) 4 1 4 lim 1 x x B x 4 4 4 lim 1 4 x x x x x x 3 1 4 4 4 4 lim 4 4 x x x x 1 7) 2 2 2 lim 1 x x x B x x 2 2 2 2 lim 2 2 x x x x x x x x x 2 2 2 lim 2 x x x x x 2 Bài 3. 1) 0 9 16 lim x x x L x 9 lim x x x x 0 9 16 lim x x x x x x 1 lim 9 16 x x x 7 24 2) 1 2 5 lim 1 x x x L x 2 2 lim x x x x 1 2 2 2 lim 1 x x x x x x 2 lim 2 2 x x x 4 3) 3 2 2 lim 3 x x x L x 2 6 2 lim 3 x x x x 3 6 2 2 6 2 lim 3 x x x x x x 3 2 6 2 lim 3 x x x x x x 3 2 lim 6 2 x x x 5 4) 2 2 4 2 2 2 lim 2 x x x x x x x x x 2 2 2 2 2 2 lim 2 x x x x x x x x 2 2 lim 2 2 x x x x x x 5) 6 5 84 lim 6 x x x x L x 5 3 16 96 lim 6 x x x x x x 2 2 lim 2 x x x x L x 2 2 2 lim 2 x x x x x (15)6 5 3 16 lim 6 x x x x x 2 5 16 2 3 lim 6 x x x x x x 6 10 lim 16 2 3 x x x 74 6) 0 1 lim x x x L x 2 0 24 10 1 lim x x x x 2 0 24 10 1 lim 24 10 1 x x x x x x 0 24 10 lim 24 10 1 x x x x x x 24 10 lim 24 10 1 x x x x 5 7) 2 1 4 3 lim 2 x x x x L x x 2 4 2 lim 1 x x x x x x x 2 2 1 4 2 lim 1 1 x x x x x x x x x x x 1 1 lim 2 x x x x x 5 8) 2 1 3 2 lim 2 x x x x L x x 4 2 lim 1 x x x x x x 2 2 4 16 48 14 49 2 2 7 lim 1 x x x x x x x x x x x 2 2 1 2 2 7 lim 1 x x x x x x x x 1 1 lim 7 2 x x x x x 17 16 9) 2 0 4 lim x x x L x 4 lim x x x x x x 2 2 4 4 6 4 lim x x x x x x x x x x x x 2 2 2 lim x x x x x x x x 0 1 lim 2 4 x x x x x (16)Page 16 10) 2 6 lim 1 x x x x L x 2 2 2 lim 1 x x x x x 2 2 6 4 2 6 lim 1 x x x x x x x x 2 2 1 2 6 lim 1 x x x x x x 1 1 lim 6 x x x 11 Bài 4. 1) 3 2 4 lim 2 x x L x 3 4 lim 2 16 4 x x x x x 3 4 lim 16 4 x x x 1 2) 3 0 1 lim x x L x 3 3 1 lim 1 1 x x x x x 3 1 lim 1 1 x x x 1 3) 3 3 1 lim 3 x x L x 2 3 2 3 2 3 9 lim 3 x x x x x 2 3 2 3 2 3 3 lim 1 x x x x 1 4) 3 1 7 lim 1 x x L x 3 3 1 lim 1 7 1 x x x x x x 2 1 3 3 1 lim 7 x x x x 1 5) 3 8 2 lim 2 x x L x 3 8 8 2 lim 2 16 2 x x x x x x 8 3 2 lim 2 x x x x 5 12 6) 3 1 lim 2 x x L x 3 1 2 3 3 1 lim 1 2 x x x x x x x 2 3 3 3 1 2 lim 1 x x x x x 1 7) 3 2 10 lim 3 x x x L x x 3 1 10 2 lim 1 x x x (17)3 2 3 3 3 1 2 1 10 2 10 lim 1 x x x x x x x 2 2 3 3 3 1 2 1 1 10 2 10 lim 1 x x x x x x x x x 2 2 3 3 3 1 2 1 10 2 10 lim 2 x x x x x x 3 8) 3 2 8 11 lim 3 x x x L x x 3 2 8 11 lim 3 x x x x 2 7 lim 3 x x x x 2 3 3 8 11 27 lim 1 11 11 x x x x x x 7 lim 1 x x x x x 2 3 3 8 lim 1 11 11 x x x x 1 lim 1 x x x 8 27 54 9) 3 3 1 1 7 lim lim lim 1 1 x x x x x x x L x x x 3 1 3 3 3 7 lim lim 1 1 7 x x x x x x x x x 3 1 3 3 3 3 1 lim lim 1 1 7 x x x x x x x x x 2 2 1 3 3 3 3 1 1 1 lim lim 4 3 7 x x x x x x x x 10) 3 0 0 2 2 lim lim lim x x x x x x x L x x x 0 3 3 4 8 lim lim 2 8 2 8 4 x x x x x x x x x 2 0 3 3 4 1 11 lim lim 12 12 2 8 2 8 4 x x x x x (18)Page 18 11) 2 3 2 2 2 2 2 11 11 lim lim lim 4 4 x x x x x x x x x L x x x 2 2 2 2 2 2 3 2 11 27 lim lim 4 4 11 11 x x x x x x x x x x x x 2 2 2 2 2 3 2 lim lim 4 4 11 11 x x x x x x x x x x x x 2 2 2 3 2 lim lim 2 2 11 11 x x x x x x x x x x 1 9 24 72 12) 3 2 0 4 2 4 4 lim lim x x x x x x x L x x x x 3 2 0 0 3 3 0 3 3 4 2 4 4 lim lim 4 8 4 lim lim 1 1 8 4) 4 lim lim 1 1 8 1 1 2 x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Bài 5. 1) 0 1 1 lim lim 1 1 n x x n n n n n ax ax F x x ax ax ax 1 0 lim 1 n n 1 x n n n a a n ax ax ax 2) 0 1 1 1 lim lim n m n m x x ax bx ax bx F x x 0 1 1 lim lim n m x x ax bx a b x x n m 3) 0 1 1 1 lim lim 1 1 n n m m x x ax ax F x bx bx (19)Xét 0 1 lim ; n x ax a A x n 1 lim m x bx b B x m 1 a am F b n bn m 4) 0 1 1 1 lim lim 1 1 n m n m x x ax bx ax bx F x x 0 1 1 lim lim 1 1 n m x x ax bx x x 0 1 1 lim lim 1 1 n m x x ax x bx x x x x x Ta có 0 1 lim n x ax a A x n 0 1 lim m x bx b B x m 0 1 lim lim lim 1 1 1 x x x x x x C x x x .2 2 a b a b F n m n mDạng Giới hạn hàm số khix→  Phương pháp giải: - Đối với dạng đa thức không căn, ta rút bậc cao áp dụng công thức khix→ + 1 lim k x→+x = + 2 lim 2 k x khi k l x khi k l →− + =  = − = +  3 lim k x c x →+ = (c số) - Đối với dạng phân sốkhông căn, ta làm tương tựnhư giới hạn dãy số, tức rút bậc cao tử và mẫu, sau áp dụng cơng thức trên - Ngồi việc đưa khỏi bậc chẵn cần có trị tuyệt đối, học sinh cần phân biệt đưa căn, liên hợp Phương pháp suy luận tương tựnhư giới hạn dãy số, cần phân biệt khix→ +hoặcx→ −  VÍ DỤ Ví dụ 1. Tính giới hạn lim( ) x A x x x →+ = − − + + Đs: Lời giải 3 lim (20)Page 20 Ví dụ 2. Tính giới hạn 3 2 3 lim 2 6 x x x B x x Đs: 1 Lời giải 3 2 2 3 3 3 3 3 1 1 1 1 0 lim lim 2 2 6 0 6 6 x x x x x x x B x x x x x Ví dụ Tính giới hạn lim x C x x x Đs: Lời giải 2 2 1 1 lim lim x x C x x x x x x x x →− →−         =   + + + =   + + +              2 1 1 lim lim x→− x x x x x→− x x x        = −  + + + =   − + + = −            (Vì lim x x 1 lim 2 1 x x x ) BÀI TẬP ÁP DỤNG Bài 1. Tính giới hạn sau: 1) lim x A x x Đs: 2) lim x A x x Đs: 3) lim x A x x Đs: 4) lim x A x x Đs: 5) lim x A x x Đs: Bài Tính giới hạn sau: 1) lim 2 x x B x Đs: B 2) lim x x B x Đs: B 3) 4 4 2 15 lim 1 x x x B x Đs: B 4) 3 3 2 lim 1 x x x B x x Đs: B 5) 2 3 lim 2 x x x B (21)6) 3 2 lim 3 x x x B x x Đs: 2 B 7) 3 7 4 lim 2 x x x B x Đs: B 8) 20 30 50 2 3 lim 1 x x x B x Đs: 30 3 B 9) 2 3 lim 4 x x x B x Đs: B 10) 3 2 lim x x x B x Đs: B Bài 3. Tính giới hạn sau: 1) lim 10 x C x x x Đs: 17 2 2) 4 2 lim 1 x x x C x Đs: 3) lim 4 13 x C x x x Đs: 14 4) lim x C x x x Đs: 2 5) lim x C x x Đs: 6) lim 2021 x C x x x Đs: 2019 7) 2 lim x C x x x Đs: 2 8) 2 2 lim 5 x x C x x Đs: -2 9) 2 lim x C x x x Đs: 4 10) 2 2 lim 2 x x x x C x Đs: 1 11) lim 1 C x x x Đs: (22)Page 22 12) 2 lim 10 x x x x C x Đs: 13) 2 lim 21 13 x C x x x x Đs: 2 14) 2 2 2 4 lim 2 x x x x x C x x x Đs: 15) lim 4 x C x x x Đs: 16) lim 3 x x C x x x Đs: -1 17) lim 16 x C x x x Đs: 43 8 18) 3 5 2 lim 3 x x x C x x x Đs: 19) lim x C x x x Đs: 2 Bài 4. Tính giới hạn sau: 1) 3 5 2 lim 3 x x x x x x Đs: 2) 2 2 lim 5 x x x x Đs: 3) 2 2 2 lim 4 1 x x x x x x Đs: 4) 4 2 2 lim 5 x x x x x x x Đs: 2 2 5) 2 2 2 lim 1 10 x x x x x x x Đs: 3 6) 3 2 3 1 lim 6 x x x x Đs: 1 7) 2 2 2 lim 5 x x x x x Đs: (23)8) 2 2 4 lim 9 x x x x x x x Đs: 4 9) 2 2 lim x x x x x x Đs: 10) 2 8 lim 6 x x x x x Đs: 11) 2 2 1 lim 3 x x x x x x Đs: 12) lim 2 x x x x Đs: Bài 5. Tính giới hạn sau: 2 1) lim x x x x Đs: 2 2) lim x x x x Đs: 3) lim 2 x x x Đs: 2 4) lim x x x x Đs: 1 2 5) lim x x x x Đs: 2 6) lim x x x x Đs: 5 3 7) lim 27 x x x x Đs: 1 27 2 8) lim x x x x Đs: 1 2 9) lim 4 x x x x Đs: 4 2 10) lim x x x x Đs: 3 2 11) lim 4 x x x x Đs: 19 2 (24)Page 24 3 2 4 13) lim 2 x x x x x x x Đs: 16 9 3 14) lim x x x Đs: LỜI GIẢI Bài 1. 1) 3 3 lim x A x x x , (vì 3 lim x x 3 lim 1 x x x ) 2) 3 3 3 lim , ì lim lim 1 x x x A x v x v x x x x 3) 4 2 4 2 lim , ì lim lim 1 x x x A x v x v x x x x 4) 4 2 4 2 3 lim , ì lim lim 1 x x x A x v x v x x x x 5) 4 2 4 1 6 lim , ì lim lim 1 x x x A x v x v x x x x Bài 2. 1) 1 1 8 8 1 8 lim lim lim 1 2 2 2 x x x x x x x B x x x x 2) 2 2 1 1 2 lim lim lim 1 1 1 1 x x x x x x x B x x x x 3) 4 4 4 4 4 4 7 15 7 15 2 2 2 15 0 lim lim lim 1 1 1 1 x x x x x x x x x x B x x x x 4) 3 3 3 3 3 3 3 4 2 2 0 lim lim lim 1 1 1 0 1 x x x x x x x x x x B x x x x x x x 5) 2 3 lim 2 x x x B x 3 2 3 3 3 3 7 0 0 lim lim 1 2 x x x x x x x x x x x x 6) 3 2 lim 3 x x x B x x 2 2 2 lim 3 x x x (25)3 3 2 4 2 lim lim 4 3 3 x x x x x x x x x x 2 3 7) 3 7 4 lim 2 x x x B x 3 3 7 3 4 4 lim 3 2 x x x x 8) 20 30 50 2 3 lim 1 x x x B x 20 30 20 30 30 50 50 3 2 2 3 lim 2 1 2 x x x x 9) 2 3 lim 4 x x x B x 2 2 2 1 1 3 3 3 lim lim , 4 1 x x x x x x x x x x x 2 1 3 ì lim lim 4 x x x x v x v x 10) 3 2 lim x x x B x 3 2 2 3 2 2 2 3 2 2 lim lim , 5 1 x x x x x x x x x x x 2 2 2 2 ì lim lim 5 x x x x v x v x Bài 3. 1) lim 10 x C x x x 10 lim x x x x 2 3 10 lim 3 x x x x x 3 10 lim 3 x x x x x 2 2 10 lim 3 1 x x x x 3 17 10 (26)Page 26 2) 4 2 lim 1 x x x C x 2 2 4 1 1 2 lim lim 1 2 x x x x x x x x x x x , 2 1 2 ì lim lim 1 2 x x x x v x v x 3) lim 4 13 x C x x x 13 lim 4 x x x x 2 2 4 4 13 lim 4 x x x x x x x 2 4 13 lim 4 4 x x x x x x 2 1 13 lim 4 4 x x x x x x x 1 13 lim 4 4 x x x x x x x 2 1 13 lim 14 4 4 x x x x 4) lim x C x x x lim 1 lim 1 x x x x x x x 1 1 1 5 lim lim 2 1 1 1 x x x x x x 5) lim x C x x lim 12 lim 12 x x x x x x x , 2 1 ì lim 2 lim x x v v x x 6) lim 2021 x C x x x 2021 lim 2021 lim x x x x x x x 4 1 4 2021 lim 2021 lim 2021 2019 2 4 1 1 x x x x x x (27)7) 2 2 1 lim lim 1 x x x C x x x x x x 2 2 1 lim lim 1 1 1 1 1 1 lim 2 1 1 x x x x x x x x x x x x x x x x 8) 2 2 lim 5 x x C x x 2 2 2 lim lim 1 5 0 1 x x x x x x x x x x x 9) 2 lim x C x x x lim 32 14 2 x x x x x 2 2 2 4 3 1 4 3 lim lim 4 3 4 x x x x x x x x x x 10) 2 2 lim 2 x x x x C x 1 1 lim 2 x x x x x 1 1 1 1 2 1 lim lim 3 3 2 2 2 x x x x x x x x 11) lim 1 x C x x x lim 1 12 x x x x x 2 1 1 lim 1 x x x x 2 2 1 1 1 lim 1 1 x x x x x x 2 1 1 1 lim 2 1 1 x x x x (28)Page 28 12) 2 lim 10 x x x x C x 1 1 1 lim lim 10 10 1 x x x x x x x x x x 13) 2 lim 21 13 x C x x x x lim 212 132 x x x x x x x 2 2 9 21 13 4 lim 9 21 13 4 x x x x x x x x x x 2 34 2 lim 2 2 9 21 13 4 x x x x x x 14) 2 2 4 lim 2 x x x x x C x x x 3 2 3 2 2 2 4 7 4 lim lim 2 1 3 2 x x x x x x x x x x x x x x x x x x x 15) lim 4 x C x x x lim 4 12 x x x x x 2 2 4 4 3 lim 4 4 x x x x x x 1 4 3 lim 4 4 x x x x 16) lim 3 x x C x x x 2 3 2 3 1 1 1 lim lim 1 5 0 1 x x x x x x x x x x x x 17) lim 16 x C x x x 3 5 lim 16 lim 16 3 16 16 3 3 43 5 lim lim 5 4 8 3 16 16 x x x x x x x x x x x (29)18) 3 5 2 lim 3 x x x C x x x 3 2 2 5 3 3 1 1 2 2 lim lim 1 1 1 x x x x x x x x x x x 19) lim x C x x x lim 1 12 x x x x x 2 1 lim 1 x x x x 2 2 1 1 3 lim 1 1 x x x x x x 2 1 1 3 lim 2 1 1 x x x x Bài 1) 3 5 2 2 1 2 lim lim lim 1 3 x x x x x x x x x x x x x x x x x x x x 2) 2 2 3 2 3 lim lim lim 1 5 5 1 x x x x x x x x x x x x x 3) 2 2 2 2 2 1 2 1 1 2 1 lim lim lim 2 1 1 4 1 4 x x x x x x x x x x x x x x x x x x x x 4) 2 4 2 2 2 2 1 3 2 2 lim lim lim 5 5 2 2 x x x x x x x x x x x x x x x x x x 5) 2 2 2 2 2 1 1 2 4 2 lim lim lim 3 1 10 1 10 1 10 1 9 x x x x x x x x x x x x x x x x x x x x x x x 6) 3 2 2 3 2 3 1 1 3 8 3 1 lim lim lim x x x x x x x x (30)Page 30 7) 2 2 1 3 2 1 2 1 2 lim lim lim 1 5 5 5 x x x x x x x x x x x x x x x 8) 2 2 2 2 2 3 4 4 1 lim lim lim 4 1 3 9 9 5 3 9 5 x x x x x x x x x x x x x x x x x x x x x x 9) 2 1 2 2 lim lim lim 2 1 1 1 1 x x x x x x x x x x x x x x x x 10) 2 2 3 8 lim lim lim 1 3 6 6 x x x x x x x x x x x x x x x 11) 2 2 2 2 2 1 1 7 1 lim lim lim 3 3 3 1 5 x x x x x x x x x x x x x x x x x x x x x 12) 2 1 2 2 2 lim lim lim 1 1 1 x x x x x x x x x x x x x x Bài 2 1 1) lim lim lim 1 x x x x x x x x x x x 1 ì lim lim 1 x x V x v x 2 2 2 4 2) lim lim lim 4 1 x x x x x x x x x x x x x x x 2 3) lim 2 lim lim 2 2 1 x x x x x x x x x x x x 4 1 ì lim lim 2 2 1 x x V v x x x (31)2 2 2 2 1 1 4) lim lim lim 1 1 1 x x x x x x x x x x x x x x x x x 2 1 1 lim 2 1 1 x x x x 2 2 2 2 4 5) lim lim lim 4 4 1 2 x x x x x x x x x x x x x x x x 2 3 lim 4 1 x x x x x 2 2 2 2 3 5 6) lim lim lim 3 3 1 x x x x x x x x x x x x x x x x x 2 4 5 lim 2 3 1 x x x x x 3 3 2 3 3 2 27 27 7) lim 27 lim 27 27 x x x x x x x x x x x x x x 2 2 2 3 23 3 3 1 lim lim 27 1 1 27 27 27 27 x x x x x x x x x x 2 2 2 2 4 2 8) lim lim lim 2 2 2 x x x x x x x x x x x x x x x x x 2 1 1 lim 2 2 x x x x 2 2 2 2 2 4 16 9) lim 4 lim lim 2 4 3 4 x x x x x x x x x x (32)Page 32 2 6 16 16 lim lim 4 3 2 4 x x x x x x x x x x x 4 4 2 4 2 2 2 4 10) lim lim lim 3 4 4 x x x x x x x x x x x x x x x x x 2 2 1 3 lim 4 3 4 x x x x 2 2 2 2 4 19 15 11) lim 4 lim lim 3 4 4 x x x x x x x x x x x x x x x x x 2 15 19 19 lim 4 3 4 x x x x x 2 2 2 2 4 16 12) lim 4 lim lim 4 4 4 x x x x x x x x x x x x x x x x x 2 8 16 lim 4 4 x x x x x 13) 3 3 2 2 2 2 3 2 3 3 2 3 4 4 lim lim 4 2 4 4 x x x x x x x x x x x x x x x x x x x x x x x 2 3 3 2 4 16 lim 3 4 4 1 1 x x x x 14) 3 3 2 2 3 3 8 lim lim 8 x x x x x x (33)2 2 3 3 12 lim 8 x x x x x x x 2 2 2 3 3 6 12 lim 1 1 8 x x x x x x xDạng Giới hạn bên xx0+ xx0− Phương pháp giải: - Sử dụng định lý giới hạn hàm số Chú ý: xx0+ x x0xx0 0 x x0 x x0 x x0 −  − →    VÍ DỤ Ví dụ 1. Tính giới hạn 1 2 lim 1 x x A x + → − = − Đs: − Lời giải Vì 1 1 lim 2 lim lim 1 1 1 x x x x x x A x x x x Ví dụ 2. Tính giới hạn 2 15 lim 2 x x A x + → − = − Đs: − Lời giải Vì 2 2 lim 15 13 15 lim lim 2 2 2 x x x x x x A x x x x Ví dụ 3. Tính giới hạn 3 2 lim 3 x x A x − → − = − Đs: − Lời giải Vì ( ) ( ) 3 3 lim 2 lim lim 3 3 3 x x x x x x A x x x x − − − → → → − − = −    −  − =  = = −  −  →    −   Ví dụ 4. Tính giới hạn 2 1 lim 2 x x A x + → + = − Đs: + (34)Page 34 Vì ( ) ( ) 2 2 lim 1 lim lim 2 2 2 x x x x x x A x x x x + + + → → → + + =    +  − =  = = +  −  →    −   Ví dụ 5. Tính giới hạn ( )2 5 lim 4 x x A x − → − = − Đs: − Lời giải Vì ( ) ( ) ( ) ( ) 4 2 2 4 2 lim 5 lim lim 4 4 x x x x x x A x x x − − − → → → − − = −    −  − =  = = −  −   →  −   Ví dụ 6. Tính giới hạn ( )2 3 lim 3 x x A x − → − = − Đs: + Lời giải Vì ( ) ( ) ( ) ( ) 3 2 2 3 2 lim 3 lim lim 3 3 x x x x x x A x x x − − − → → → − − =    −  − =  = = +  −   →  −   Ví dụ 7. Tính giới hạn ( ) ( ) 2 2 lim 3 x x x A x + → − + − = + Đs: − Lời giải Ta có ( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 3 3 2 2 lim lim lim 3 3 x x x x x x x x x x x + + + → − → − → − − + + − = = − + + + Vì ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 3 lim 2 lim lim 3 3 3 x x x x x x x A x x x x + + + → − → − → − +  − = −   + −  + =  = = −  +   → −   −  +   Ví dụ 8. Tính giới hạn 2 2 1 lim 2 x A x x − →   =  −  − −   Đs: − Lời giải Ta có: ( )( ) 2 2 1 1 lim lim 2 2 x x x A x x x x − − → → +   =  − = − − − + (35)Vì ( ) ( )( ) ( )( ) 2 2 2 lim 1 lim 2 lim 2 2 2 x x x x x x A x x x x x x − − − → → → − + =     − + =  =  − = −     − −   →    − +   Ví dụ 9. Tính giới hạn 2 2 2 lim 2 x x B x x − → − = − + Đs: 1 − Lời giảix→2−   − = −x 2 x x Do ( )( ) 2 2 1 lim lim 2 2 x x x B x x x − − → → − − = = = − − − − Ví dụ 10.Tính giới hạn 3 3 lim 5 15 x x B x + → − = − Đs: 1 Lời giảix→3+    − = −x x x Do ( ) 3 3 1 lim lim 5 5 x x x B x − − → → − = = = −  BÀI TẬP ÁP DỤNG Bài 1. Tính giới hạn sau: 1) 3 1 1 lim 2 x x A x x − → − = + − Đs: 1 − 2) 2 2 lim 2 x x B x → − = − Đs: Không tồn 3) 2 3 9 lim 3 x x C x → − = − Đs: Khơng tờn Bài 2. Tính giới hạn sau: 1) 2 2 2 lim 2 x x x x x C x x − → − + − + = − + Đs: 7 2) 2 2 lim 1 x x C x − → − = − − Đs: −2 3) 2 2 7 12 lim 9 x x x D x − → − + = − Đs: (36)Page 36 5) 2 1 1 lim x x x D x x − → − + − = − Đs: 6) ( ) 3 2 1 5 lim 2 x x D x x x + → + = − + − Đs: 7) 3 3 lim 5 x x x D x x − → − + = − + Đs: 3 Bài 3. 1)Tính giới hạn ( ) 1 lim x C f x → = với ( ) 4 3 5 3 x x x khi x f x x x khi x  − −   =  −   Đs:−2 2) Tính giới hạn ( ) 1 lim x C f x → = với ( ) 2 3 1 x khi x f x x khi x −   =  − +   Đs:−2 3) Tính giới hạn ( ) 2 lim x C f x →− = với ( ) 3 2 10 x khi x f x x x khi x −   −  = +  +  −  Đs:8 Bài 4. Tìm m để hàm số ( ) 3 2 1 1 1 x khi x f x x mx x m khi x  +  −  = +  − +  −  có giới hạn x= −1 Đs: m=1 m= −2  LỜI GIẢI Bài 1. 1) 3 1 1 lim 2 x x A x x − → − = + − Vì x→    − = − −1− x x (x ) Do ( ) ( )( ) 1 1 1 lim lim 2 1 2 x x x A x x x x x − − → → − − − = = = − + + − + + 2) 2 2 lim 2 x x B x → − = − +) Vì x→2−   − = − −x x (x 2) nên ( ) ( ) 2 2 lim lim 1 2 x x x x − − → → − − = − = − − +) Vì x→2+   − = −x x x nên 2 2 lim lim 1 x x x x − − → → − = = − Suy 2 2 lim lim 2 x x x x x x − + → → − −  − − nên không tồn giới hạn 2 lim 2 x x B x → − = (37)3) 2 3 9 lim 3 x x C x → − = − Ta có 3 3 lim 3 x x x C x → − + = − Do đó: +) ( ) 2 3 3 9 3 3 lim lim lim 3 x x x x x x x x x + + + → → → − − + = = + = − − +) ( ) ( ) 2 3 3 9 lim lim lim 3 x x x x x x x x x − − − → → → − − − + = = − + = − − − Suy giới hạn 2 3 9 lim 3 x x C x → − = − không tồn Bài 2. 1) 2 2 2 lim 2 x x x x x C x x − → − + − + = − + Vì x→  −   − = − −1− x x (x 1) Do ( ) ( ) ( ) ( )( ) 2 1 1 2 1 3 lim lim lim 1 1 x x x x x x x x x x x C x x x x x − − − → → → − − − + − + − − = = = − − − + + ( )( ) ( )( ) 1 1 4 lim lim 4 2 1 x x x x x x x x x x − − → → − + + = = = + + − + + 2) 2 2 lim 1 x x C x − → − = − − Vì x→2−  −   − = − −x x (x 2) Do đó: ( )( ) ( ) ( ) 2 2 1 lim lim 1 1 x x x x C x x − − → → − − − +   = = − − + = − − − 3) 2 2 7 12 lim 9 x x x D x − → − + = − Ta có ( )( ) ( )( ) 3 3 3 3 4 4 1 lim lim lim 3 3 3 x x x x x x x x D x x x x x − − − → → → − − − − − = = = = − + + − + 4) 2 2 5 lim 4 x x x D x − → − + = − Ta có ( )( ) ( )( ) 2 2 2 2 3 3 1 lim lim lim 2 2 2 2 x x x x x x x x D x x x x x − − − → → → − − − − − = = = = − + + (38)Page 38 5) 2 1 1 lim x x x D x x − → − + − = − Ta có ( ) ( ) ( )2 1 1 1 1 1 lim lim lim 1 x x x x x x x x D x x x x x − − − → → → − − − − − − − − = = = = − − 6) ( ) 3 2 1 5 lim 2 x x D x x x + → + = − + − Ta có ( ) ( ) ( )( ) ( )( ) 2 2 1 1 5 lim lim 3 1 3 x x x x x x D x x x x x + + → →  − +   − +    = − = − = + +  − + +      7) 3 3 lim 5 x x x D x x − → − + = − + Ta có ( ) ( ) ( )( ) (( ))( ) 2 1 1 1 2 lim lim lim 1 4 x x x x x x x x D x x x x x − − − → → → − + − + + = = = = − − − − − Bài 1)Ta có: +) ( ) ( ) 1 lim lim x x f x x x − − → = → − = − +) ( ) ( ) 1 lim lim 6 x x f x x x x + + → = → − − = − − = − +) Vì ( ) ( ) 1 lim lim x x f x f x − + → = → = − nên hàm số f x( ) có giới hạn x=1 ( ) 1 lim xf x = − 2) Ta có: +) ( ) ( ) 1 lim lim x x f x x − − → = → − = − +) ( ) ( ) 1 lim lim 2 x x f x x + + → = → − + = − +) Vì ( ) ( ) 1 lim lim x x f x f x − + → = → = − nên C=limx→1 f x( )= −2 3) Ta có: +) ( )2 ( ) ( )2 3 lim lim 1 x x x f x x − − → − → − − = = + +) ( )2 ( ) ( )2 ( ) lim lim 10 x x f x x + + → − = → − + = +)Vì ( )2 ( ) ( )2 ( ) lim lim x x f x f x − + → − = → − = nên C=xlim→−2 f x( )=8 (39)+) ( ) ( ) ( ) ( ) ( ) 3 2 1 1 1 lim lim lim 1 x x x x f x x x x − − − → − → − → − + = = − + = + +) ( ) ( ) ( ) ( ) 2 2 1 lim lim x x f x mx x m m m + + → − = → − − + = + + +) Để hàm số có giới hạn x= −1 2 3 2 m m m m m m =  = + +  + − =   = −  Dạng Giới hạn hàm số lượng giác Phương pháp giải: - Sử dụng định lý giới hạn hàm số - Sử dụng công thức biến đổi lượng giác - Lưu ý: 0 sin lim x x x VÍ DỤ Ví dụ 1. Tính giới hạn 2 6 2sin lim 4 cos x x A x  → − = − Đs: 1 A= − Lời giải Ta có: ( ) 2 2 6 6 2sin 2sin 2sin 1 lim lim lim lim 4 cos sin 4sin 2sin x x x x x x x A x x x x     → → → → − − − − = = = = = − − − − − + Ví dụ 2. Tính giới hạn 2 4 2 sin lim 2 cos x x A x  → − = − Đs: 1 A= − Lời giải Ta có: ( ) 2 2 4 4 2 sin sin sin 1 lim lim lim lim 2 cos sin 1 2sin sin x x x x x x x A x x x x     → → → → − − − − = = = = = − − − − − + Ví dụ 3. Tính giới hạn 0 cos lim sin x x A x → − = Đs: A=0 Lời giải Ta có: 2 2 0 cos cos sin cos sin lim lim sin 2sin cos x x x x x x x A x x x → → − − − − = = 2 0 2sin sin lim lim 2sin cos cos x x x x x x x → → − − = = = Ví dụ 4. Tính giới hạn 0 1 sin cos lim 1 sin cos x x x A x x → − − = (40)Page 40 Ta có: ( ) ( ) 2 2 0 1 2sin cos cos sin sin cos lim lim 1 sin cos 2sin cos cos sin x x x x x x x x A x x x x x x → → − − − − − = = + − + − − ( ) ( ) 2 0 0 2sin sin cos 2sin 2sin cos sin cos lim lim lim 2sin 2sin cos 2sin sin cos sin cos x x x x x x x x x x x x x x x x x x x → → → − − − = = = = − + + +  BÀI TẬP ÁP DỤNG Bài 1. Tính giới hạn sau: 1) 0 1 sin cos lim 1 sin cos x x x A x x → + − = − − Đs: A= −1 2) sin lim 1 sin cos x x A x x → = − − Đs: A= −1 3) 0 sin sin lim sin x x x A x → − = Đs: A=2 4) 0 sin sin lim sin x x x A x → − = Đs: A=2 5) 0 1 cos lim sin x x A x → − = Đs: A=0 6) 3 cos cos 2 lim sin x x x A x  → + + = Đs: 3 A= 7) 2 1 sin cos lim cos x x x A x  → + + = Đs: A=2 Bài 2. Tính giới hạn sau: 1) 0 1 cos lim 1 cos x ax B bx → − = − Đs: 2 a B b   =    2) 0 sin lim x x B x → = Đs: B=5 3) 3 0 sin sin sin lim 45 x x x x B x → = Đs: 3 B= 4) 2 0 1 cos lim x x B x → − = Đs: 2 B= 5) 0 1 cos lim 1 cos x x B x → − = − Đs: 25 B= 6) 2 0 1 cosa lim x x B x → − = Đs: 2 2 a B= 7) 2 0 1 cos lim sin x x B x x → − = Đs: B=4 8) 3 0 sin tan lim x x x B x → − = Đs: B= − 9) 3 0 tan sin lim sin x x x B x → − = Đs: 2 B= 10) 3 0 1 cos lim sin x x B x x → − = Đs: 2 B= Bài 3. Tính giới hạn sau: 1) ( ) 2 4 cos8 sin lim 3 x x x B x → − = Đs: B= −48 2) 0 1 lim sin x x B x → − + = Đs: 2 B= − 3) 2 0 1 cos cos lim x x B x → − = Đs: 2 B= 4) 3 1 cos lim tan x x B x → − = Đs: 6 (41)5) 3 tanx lim 2sin x B x  → − = − Đs: 1 B= 6) 3 0 1 tan sin lim x x x B x → + − + = Đs: 4 B= 7) ( )2 0 1 cos lim 1 x x B x → − = − − Đs: B=2 8) 2 1 cos lim x x x B x → + − = Đs: B=1 9) 0 1 sin lim 3 x x x B x x → − + + = + − − Đs: B=0 10) 3 0 2 1 lim sin x x x B x → + − + = Đs: Bài Tính giới hạn sau: 1) 4 lim tan tan x C x x   →    =   −      Đs: 1 C= 2) ( )2 1 cos lim x x C x   → + = − Đs: 1 C= 3) lim sin2 ( 1) x x C x x  → − = − + Đs: 1 C= − 4) limsin sin x a x a C x a → − = − Đs: C=cos a LỜI GIẢI Bài 1. 1) ( ) ( ) 2 2 0 1 2sin cos cos sin sin cos lim lim 1 sin cos 2sin cos cos sin x x x x x x x x A x x x x x x → → + − − + − = = − − − − − ( ) ( ) 2 0 0 2sin sin cos 2sin 2sin cos sin cos lim lim lim 2sin 2sin cos 2sin sin cos sin cos x x x x x x x x x x x x x x x x x x x → → → + + + = = = = − − − − 2) ( 2 ) 0 sin 2sin cos lim lim 1 sin cos 2sin cos cos sin x x x x x A x x x x x x → → = = − − − − − ( ) 2 0 0 2sin cos 2sin cos cos lim lim lim 2sin 2sin cos 2sin sin cos sin cos x x x x x x x x x x x x x x x x → → → = = = = − − − − 3) 0 0 sin sin 2cos sin lim lim lim 2cos sin sin x x x x x x x A x x x → → → − = = = = 4) 0 0 sin sin 2cos sin lim lim lim 2cos sin sin x x x x x x x A x x x → → → − (42)Page 42 5) 2 0 0 2sin sin 1 cos 2 2 lim lim lim sin 2sin cos cos 2 2 x x x x x x A x x x x → → → − = = = = 6) ( ) 3 2 3 3 4 cos 3cos cos sin cos cos 2 lim lim sin 3sin 4sin x x x x x x x x A x x x   → → − + − + + + = = − ( ) ( ( ) ) 2 3 2 3 cos cos cos cos 3cos cos lim lim sin 4sin sin cos x x x x x x x x x x x x   → → − + − + = =   −  − −  ( ) ( )( ) ( )( ) (( )) 2 2 3 3 cos cos cos 2 cos 3 cos 1 cos 2 cos 3 2 3 lim lim lim sin cos cos sin cos sin cos x x x x x x x x x x x x x x x x x    → → →  + −  + − +   = = = = − + +  −    7) 2 2 1 sin cos 2 cos 2sin cos lim lim cos cos x x x x x x x A x x   → → + + + = = ( ) ( ) 2 2 cos cos sin lim lim cos sin cos x x x x x x x x   → → + = = + = Bài 1) 2 2 0 2 2sin sin 1 cos 2 2 2 lim lim lim 1 cos 2sin sin 2 2 x x x ax ax bx ax a a A bx ax bx bx b b → → →     −   = = =   =  −       (Vì 0 sin lim 2 x ax ax → = 2 lim sin x bx bx → = ) 2) 0 sin sin lim lim 5 5 x x x x B x x → →   = =  =   (Vì sin lim 5 x x x → = ). 3) 3 0 sin sin sin sin sin sin 1 lim lim 45 3 x x x x x x x x B x x x x → →   = =  =   (Vì 0 sin lim 5 x x x → = , sin lim 3 x x x → = , sin lim x x x → = ) 4) 2 2 0 2sin 1 cos 2 lim lim 2 x x x x B x x → → − = = =       , (vì 2 2 sin lim 2 x x x →   =     . 5) 2 2 2 0 2 2 5 5 sin . 2sin 1 cos 2 2 25 25 lim lim lim 3 1 cos 2sin 9 .sin 2 2 2 x x x x x x x B x x x x → → →         −     = = = =   −      (43)(Vì 2 2 5 sin 2 lim 5 x x x →   =     2 0 2 3 lim 3 sin 2 x x x →       = ) 6) 2 2 2 0 2sin 1 cosa 2 lim lim 4 2 x x ax x a a B x ax → →     −   = = =           , (vì 2 2 sin lim 2 x ax ax →   =     ). 7) 2 2 2 0 0 sin 4sin cos sin lim lim lim cos .sin sin x x x x x x x B x x x x x x → → →   = = =  =   , (vì sinx lim xx = ). 8) 3 3 3 0 0 sin sin sin tan cos sin cos sin lim lim lim cos x x x x x x x x x x x B x x x x → → → − − − = = = ( ) 2 0 sin sin cos 2sin 2 lim lim cos cos 2 x x x x x x x x x x x → →     − − − −  − = = =            (vì 0 sinx lim xx = 2 2 sin lim 2 x x x →       =       ) 9) 3 3 3 0 0 sin sin tan sin cos sin sin cos lim lim lim sin sin sin x cos x x x x x x x x x x x B x x x → → → − − − = = = 2 2 0 2 2 2 2sin 1 cos 2 1 lim lim lim sin x cos 4.sin cos cos cos cos 2 2 x x x x x x x x x x x → → → − = = = = 10) ( )( ) ( ) 2 2 0 2sin cos cos 1 cos cos cos 2 lim lim sin 2 sin cos 2 x x x x x x x x B x x x x x → → + + − + + = = 2 0 sin 1 cos cos lim 2 cos 2 x x x x x x →    + +  =  =     , (vì 0 sin lim 2 x x x → = ) ( ) ( ) (44)Page 44 = 2 0 sin sin 96 lim 48 4 cos8 x x x x x x →     −  = −      +      2) 0 1 2 1 lim lim sin sin 2 x x x x B x x x → → − +  −  = =  = − + +   3) ( ) ( ( )) 2 2 2 2 2 0 0 1 cos sin 1 cos cos cos cos lim lim lim 1 cos cos cos cos x x x x x x x x B x x x x x x x → → → − − − − = = = + + ( ) ( ) ( ) 2 2 2 2 2 2 0 2 2 0 sin cos cos 2sin sin 2sin cos lim lim 1 cos cos cos cos sinx cos lim 2 cos cos x x x x x x x x x x x x x x x x x x x x → → → + − − + = = + +   +  =   = +       4) ( ) 3 2 0 3 3 1 cos cos lim lim sin tan 1 cos cos cos x x x x B x x x x x → → − − = = + + ( ) ( ) 2 2 0 2 2 3 3 2 2 3 3 2 4sin cos cos 2 lim lim 6 2sin cos cos cos cos cos cos 2 x x x x x x x x x x x x x → → = = = + + + + 5) ( )( ) 3 2 2 2 3 2 3 4 tanx tan lim lim 2 sin sin cos tan tan 1 x x x B x x x x x   → → − − = = − − + + ( 2 )(3 ) ( )(3 ) 4 sin cos 1 cos lim lim 3 sin cos tan tan cos sin cos tan tan x x x x x x x x x x x x x x   → → − = = = − + + + + + 6) ( ) 3 3 0 1 tan sin tan sin lim lim 1 tan sin x x x x x x B x x x x → → + − + − = = + + + ( ) ( ) ( ) 2 3 0 2 0 2 sin sin sin sin cos 2 lim lim cos tan sin cos tan sin sin sin 2 lim 4 tan sin 2 x x x x x x x x x x x x x x x x x x x x x x → → → − = = + + + + + +           =   =    + + +      (45)7) ( ) ( ) ( ) 2 2 2 2 0 0 2 sin 1 sin 2 1 1 1 cos 2 2 lim lim lim 4 1 2 x x x x x x x x B x x x → → →    + −   + −  −   = = =   =   − −        8) ( ) ( ) 2 2 2 2 0 2 2 1 cos cos sin lim lim lim 1 cos cos x x x x x x x x x B x x x x x x x → → → + − + − + = = = + + + + 2 2 2 2 0 sin 1 1 = lim 2 1 cos cos x x x x x x x →   + = + =   + + + +   9) 0 0 1 sin sin lim = lim lim 3 4 x x x x x x x B x x x x x x → → → − + + − + = + + − − + − − + − − ( ) ( )( ) ( ( ) ) ( ) ( )( ) 2 0 0 2 sin lim lim 1 1 2 sin 3 4 2 lim lim 1 1 4 x x x x x x x x x x x x x x x x x x x x x x x x → → → → − + + + + + + = + − + − − + + − + + +  + + +  = +   − − − − + +   = − = 10) 3 3 0 0 2 1 1 1 1 1 lim lim lim lim sin sin sin sin x x x x x x x x x x B x x x x → → → → + − + + − + − + + − − + = = = + ( ) ( ) 2 0 3 2 2 3 2 lim lim sin 1 sin 1 1 1 x x x x x x x x x → → − = + =   + +  + + + +    Bài 4. 1) 4 lim tan tan x Cxx →    =   −      Đặt 4 t= −x  , x→  → t Khi đó: ( ) 0 0 cos lim tan ( 1) tan lim cot tan lim 2 cos t t t t C t t t t t  → → →     =   +  − = = =     2) ( )2 1 cos lim x x C x   → + = − Đặt t = −x , x→  → t Khi đó: 2 2 0 2 sin 1 cos 2 lim lim 2 t t t t C t t → → − = = = ( ) (46)Page 46 Đặt t = −x , x→  →1 t Khi đó: ( ) ( ) ( )( ) ( ) 2 0 sin sin sint lim lim lim 4 3 2 x x t x x C x x x x t t   → → → − − = = = = − − + − − − 4) limsin sin x a x a C x a → − = − Đặt t= −x a x→  →a t Khi đó: ( ) 0 2 cos sin sin sin 2 2 lim lim cos 2 t t t a t t a a C a t t → → + + − = = = C BÀI TẬP RÈN LUYỆN Bài 1. Tính giới hạn sau: 1 2 3 lim x x x x → − − − ĐS: 1 5 2 15 lim x x x x → + − − ĐS : 3 2 3 lim x x x x →− + + − ĐS: 1 4 2 2 lim x x x x → − + − ĐS: 1 2 lim x x x x →− + + − ĐS: 1 − 2 12 lim x x x x → − + − ĐS: 1 − 2 1 lim x x x x → − + − ĐS: 2 5 2 2 lim x x x x → + − − ĐS: 2 2 14 lim x x x x → + − − ĐS: 11 4 10 2 9 l im 4 x x x x → − − + ĐS: 11 2 10 lim 18 x x x x x → − − + − ĐS: 11 7 12 2 5 lim 25 x x x x → − − ĐS: 1 13 2 lim 2 10 12 x x x x → − − + ĐS: 14 2 2 lim x x x x → − − − ĐS: 4 − 15 2 lim x x x x x → − + − ĐS: 1 3 16 2 20 lim x x x x x → − + − ĐS: 1 17 2 3 10 lim x x x x x → − + − + ĐS: 18 2 3 lim x x x x x → + − − − ĐS: 19 2 lim x x x x x → − + − ĐS: 1 − 20 4 2 16 lim x x x x →− − + + ĐS: −16 21 2 lim x x x x → − − + ĐS: 12 22 3 2 lim 11 18 x x x x →− + + + ĐS: (47)23 2 lim 2 x x x x → − − + − ĐS: 2 − 24 2 lim x x x x → − − + ĐS: 12 25 2 2 lim x x x →− + − ĐS: 3 − 26 ( ) 3 1 lim x x x → + − ĐS: 27 ( ) 3 27 lim x x x → + − ĐS: 27 28 4 27 lim 2 x x x x x → − − − ĐS: 29 3 2 5 10 lim 2 x x x x x → − + − − ĐS: 30 3 2 2 lim 1 x x x x x → − + + − ĐS: −1 31 2 lim x x x x → − − − ĐS: 5 2 32 3 2 lim x x x x x x →− + + − − ĐS: 2 − 33 2 10 lim x x x x x →− − − − + ĐS: 9 11 − 34 2 1 lim x x x x x x → − − + − + ĐS:2 35 lim x x x x → − − − ĐS: 4 9 36 3 2 2 lim x x x x x → − − + − ĐS: 3 37 3 lim x x x x x → + − + − ĐS: 5 8 38 3 2 3 lim 3 x x x x x x → − − + − − ĐS: 1 − 39 2 lim x x x x x x → + − − − + ĐS: 11 40 3 2 lim x x x x x → − + − + ĐS: -1 41 2 lim 3 x x x x x x →− − − + − + ĐS: 6 19 − 42 3 1 lim x x x x → − − + ĐS: 3 43 5 lim 8 x x x x x x → − + + − − ĐS: 44 3 4 1 6 lim 9 x x x x x x → − + − + − ĐS: 2 45 lim x x x x x → + − − + ĐS: 4 − 46 3 lim x x x x x → − + − + ĐS: 1 47 2 2 lim x x x x x → − + − − ĐS: 17 4 48 4 3 1 1 lim 5 x x x x x x x → − − + − + − ĐS: 3 − 49 3 2 lim 4 13 x x x x x x x → − − − − + − ĐS: 11 17 50 3 3 1 2 lim 1 x x x x x x x →− + + + + − − ĐS: 1 51 3 2 lim 4 12 12 x x x x x x x → − − − − + − ĐS: 11 20 52 3 2 lim ( 1) x x x x → − + − ĐS: 1 53 4 3 2 2 4 lim 3 14 20 x x x x x x x x →− + + − − + + + ĐS: 7 − 54 3 2 2 lim 3 x x x x x →− − + + + − ĐS: 7 55 4 5 limxx + xx+ ĐS: 56 5 5 (48)Page 48 57 2 1 1 lim 1 xx x  −   − −    ĐS: 1 2 58 1 12 lim 2 xx x  −   − −    ĐS: 1 59 2 2 2 1 lim 3 xx x x x  +   − + − +    ĐS: −2 60 2 2 26 lim 2 x x x x x →− − −  −   + −    ĐS: 7 61 2 3 1 1 lim 2 xx x x  −   + − −    ĐS: 2 9 62 (1 )(1 )(1 ) lim x x x x x → + + + − ĐS: 63 1 1 lim 1 n m x x x → − − ĐS: n m 64 1 lim ( 1) n x x nx n x → − + − − ĐS: ( 2)( 1) nn− 65 100 50 2 lim 2 x x x x x → − + − + ĐS: 66 2 1 lim 1 n x x x x n x → + + + − − ĐS: ( 1) n n+ Lời giải 1 ( )( ) 2 3 3 3 1 lim lim = lim 6 x x x x x x x x x x → → → − = − = − − + − + 2 ( )( ) ( ) 2 3 3 3 2 15 lim lim = lim 3 x x x x x x x x x x → → → − + + − = + = − − 3 2 3 3 lim 2 x x x x →− + + − 3( )( ) 3 lim 3 x x x x →− + = + − 4 2 2 3 lim 4 x x x x → − + − ( )( ) ( )( ) 2 1 lim 2 x x x x x → − − = − + 1 lim 1 x→− x = = − 1 lim 2 x x x → − = = + 5 ( )( ) ( )( ) 2 2 2 1 3 1 lim lim = lim 4 2 x x x x x x x x x x x x →− →− →− + + + + = + = − − − + − 6 ( )( ) ( )( ) 2 3 3 3 7 12 lim lim = lim 9 3 x x x x x x x x x x x x → → → − − − + = − = − − − + + 7 ( )( ) ( )( ) 2 1 1 1 1 lim lim = lim 3 4 x x x x x x x x x x x x → → → − + − = + = + − − + + 8 ( )( ) ( )( ) 2 2 2 2 6 lim lim = lim 4 2 x x x x x x x x x x x x → → → − + + − + = = − − + + 9 ( )( ) ( )( ) 2 2 2 2 2 14 11 lim lim = lim 4 2 x x x x x x x x x x x x → → → − + + − = + = − − + + 10 ( )( ) ( )( ) 2 3 3 3 9 l im lim = lim 4 3 1 x x x x x x x x x x x x → → → − + − = + = − + − − − 11 ( )( ) ( )( ) 2 2 2 2 3 10 11 lim lim = lim 4 18 9 17 x x x x x x x x x x x x x → → → − + − − = + = (49)12 ( ) ( )( ) 2 5 5 5 5 lim lim = lim 25 5 x x x x x x x x x x x x → → → − − = = − − + + 13 ( )( ) ( )( ) ( ) 2 2 2 2 4 lim lim lim 2 10 12 2 3 x x x x x x x x x x x x → → → − + − − − = = = − + − − − 14 ( )( ) ( )( ) 2 2 2 2 4 lim lim lim 2 2 3 x x x x x x x x x x x x → → → − + − − − − = = = − − − + + 15 ( )( ) ( ) 2 3 3 2 5 lim lim = lim 3 2 x x x x x x x x x x x x → → → − − − + = − = − − 16 ( )( ) ( ) 2 5 5 4 9 20 lim lim = lim 5 5 x x x x x x x x x x x x x → → → − − − + = − = − − 17 ( )( ) ( )( ) 2 3 3 3 5 lim lim = lim 3 x x x x x x x x x x x x x → → → − − − + = − = − − − − 18 ( )( ) ( )( ) 2 3 3 1 2 3 lim lim = lim 2 1 2 x x x x x x x x x x x x x → → → − + + − + = = − − − − − 19 ( )( ) ( )( ) ( ) 3 2 3 3 2 5 lim lim = lim 9 3 x x x x x x x x x x x x x x x → → → − − − − + = = − − − + − − 20 ( )( )( ) ( )( ) ( )( ) 2 4 2 2 4 2 16 lim lim = lim 16 6 4 x x x x x x x x x x x x x x →− →− →− + − + + − − = = − + + + + + 21 3 2 8 lim 5 x x x x → − − + = ( )( ) ( )( ) 2 2 2 lim 2 x x x x x x → − + + − − 22 ( )( ) ( )( ) 2 2 2 2 8 lim lim 11 18 x x x x x x x x x x →− →− + − + + = + + + + ( ) 2 2 lim 12 3 x x x x → − + + = = − 2 2 12 lim 9 x x x x →− − + = = + 23 ( )( ) ( )( ) 2 3 2 2 2 2 2 2 2 lim lim = lim 6 2 2 2 2 x x x x x x x x x x x x x x → → → − − + − − + = − + = − − − + + + + 24 ( )( ) ( )( ) 2 3 2 2 2 2 8 lim lim = lim 12 3 2 x x x x x x x x x x x x x x → → → − + + − + + = = − + − − − 25 ( )( ) ( )( ) 2 3 2 2 2 2 2 2 2 lim lim = lim 2 2 2 x x x x x x x x x x x x x →− →− →− − − + + = − + = − (50)Page 50 26 ( ) ( ) 3 3 2 2 0 0 1 3 lim lim lim 3 x x x x x x x x x x x → → → + − + + = = + + = 27 ( ) ( ) ( ) ( ) ( ) 2 2 0 0 3 3 1 27 lim lim = lim 3 27 x x x x x x x x x x x → → →  + + + +  + −     =  + + + + = 28 ( )( ) ( )( ) ( ) 2 4 3 3 3 9 27 lim lim = lim 2 3 x x x x x x x x x x x x x x x x x → → → − + + + + − = = − − − + + 29 ( )( ) ( ) 2 3 2 2 2 2 5 10 lim lim = lim 2 x x x x x x x x x x x x x → → → − − + − + − = − + = − − 30 ( )( ) ( )( ) 2 3 2 2 1 1 1 2 2 lim lim = lim 1 1 x x x x x x x x x x x x x x x → → → − − − − + + = − − = − − − + + 31 ( )( ) ( )( ) 2 3 2 2 2 2 2 2 2 lim lim = lim 4 2 2 x x x x x x x x x x x x x x → → → − + + − − = + + = − − + + 32 ( )( ) ( )( ) ( ) 3 2 2 2 1 3 2 lim lim = lim 6 3 x x x x x x x x x x x x x x x x →− →− →− + + + + + = = − − − + − − 33 ( )( ) ( )( ) 2 3 2 2 2 2 2 10 lim lim lim 6 2 3 11 x x x x x x x x x x x x x x x →− →− →− + − − − + = = = − − + + − + − + 34 ( ) ( ) ( ) ( ) 2 3 2 1 1 1 1 lim lim = lim 2 1 x x x x x x x x x x x x → → → − + − − + = + = − + − 35 ( )( ) ( )( ) ( ) 2 2 3 2 2 2 4 lim lim = lim 3 2 1 x x x x x x x x x x x x → → → − + − = + = − − − + + 36 ( )( ) ( )( ) 2 3 2 2 2 2 2 2 lim lim = lim 4 2 x x x x x x x x x x x x x → → → − − − − + = − = − − + + 37 2 3 2 1 1 3 ( 1)( 4) lim lim lim 2 ( 1)(2 3) 3 x x x x x x x x x x x x x x x → → → + − − + + = = = + − − + + + + 38 3 2 2 1 1 3 ( 1)(3 3) 3 lim lim lim 3 ( 1)(3 1) x x x x x x x x x x x x x x x x → → → − − + − − − − − = = = − − − − + + 39 3 2 2 2 2 5 ( 2)( 1) lim lim lim 11 3 ( 2)( 1) x x x x x x x x x x x x x x x x → → → + − − − + + + + = = = − + − − − 40 3 2 2 1 1 2 ( 1)(2 3) 2 lim lim lim 3 ( 2)( 1) x x x x x x x x x x x x x x x → → → − + − + − + − = = = − − + − − − 41 2 3 2 2 2 2 ( 2)( 4) lim lim lim 3 ( 2)(3 3) 3 19 x x x x x x x x x x x x x x x x →− →− →− − − + − − = = = − (51)42 3 2 4 3 1 1 1 ( 1)( 1) lim lim lim 4 ( 1)( 3) 3 x x x x x x x x x x x x x x x x x x → → → − = − − − − = − − − = − + − + − − + − − 43 3 2 4 3 3 3 5 ( 3)( 3) lim lim lim 8 ( 3)( 3) 3 x x x x x x x x x x x x x x x x x x x x → → → − + + = − − − = − − = − − − + + + + + + 44 3 2 4 3 1 1 3 3 6 (3 1)(2 1) 2 lim lim lim 9 (3 1)(3 1) 3 x x x x x x x x x x x x x x x x x x x x → → → − + − = − − + = − + = + − − + + + + + + 45 1 1 2 ( 1)( 3) lim lim lim 3 5 ( 1)( 4) x x x x x x x x x x x x x → → → + − = − + = + = − − + − − − 46 3 4 2 1 1 3 ( 2)( 1) lim lim lim 4 ( 1)( 3) x x x x x x x x x x x x x x x x x → → → − + = + − + = + = − + − + + + + + 47 5 4 2 2 2 2 ( 2)( 1) 17 lim lim lim 4 ( 2)( 2) x x x x x x x x x x x x x → → → − + − = − + = + = − − + + 48 4 2 3 2 1 1 1 ( 1)( 1) lim lim lim 5 ( 1)( 3) x x x x x x x x x x x x x x x x x x x → → → − − + = − + + + = + + = − − + − − + − − 49 3 2 3 2 3 3 2 ( 3)(2 1) 11 lim lim lim 4 13 ( 3)(4 1) 17 x x x x x x x x x x x x x x x x x x x → → → − − − − + + + + = = = − + − − − + − + 50 3 2 3 2 1 1 2 (2 1)( 1) 1 lim lim lim 1 ( 1)( 1) x x x x x x x x x x x x x x x x x →− →− →− + + + = + + + = + = + − − − + + − 51 3 2 3 2 3 3 2 ( 3)(2 1) 11 lim lim lim 4 12 12 4( 3)( 1) 4( 1) 20 x x x x x x x x x x x x x x x x x → → → − − − = − + + = + + = − + − − + + 52 3 3 2 2 3 2 2 3 2 2 1 3 2 ( 1) 1 lim lim lim ( 1) ( 1) ( 1) ( 1) x x x x x x x x x x x x → → → − + = − = = − − + + + + 53 4 2 2 3 2 2 2 2 4 (2 1)( 4) lim lim lim 3 14 20 (3 2)( 4) x x x x x x x x x x x x x x x x x x →− →− →− + + − − − + + − = = = − + + + + + + + 54 3 2 2 3 2 ( 3)(2 (3 3) 3) lim lim 3 ( )( ) x x x x x x x x x x x →− →− − + + + = + − − + − − − + 2 3 2 (3 3) 3 lim 6 x x x x →− − − + − = = − 55 4 3 4 2 1 1 5 ( 1) ( 2) lim lim lim 3 ( 1) ( 2)( 1) x x x x x x x x x x x x x x x x x x → → → − + − + − − − = = = (52)Page 52 56 5 4 2 1 5 ( 1)( 5) lim lim 1 ( 1)( 1) x x x x x x x x x x x x x x x → → + + + + − = − + + + + − − + 4 1 2 15 lim 1 x x x x x x → + + + + = = + 57 2 1 1 1 1 lim lim lim 1 ( 1)( 1) x x x x x x x x x → → → −  − = = =  − −  − + +   58 3 2 2 2 2 1 12 ( 2)( 4) lim lim lim 2 ( 2)( 4) x x x x x x x x x x x x x → → → − + +  − = = =  − −  − + + + +   59 2 2 2 2 1 2( 2) lim lim lim 3 ( 2)( 3)( 1) ( 3)( 1) x x x x x x x x x x x x x → → → −  + = = = −  − + − +  − − − − −   60 2 2 2 2 26 2( 5)( 2) 2( 5) lim lim lim 2 ( 2)( 2) 2 x x x x x x x x x x x x x →− →− →− − − − + −  − = = =  + −  − + −   61 2 3 2 2 1 1 1 ( 1)( 1) lim lim lim 2 ( 1)( 2)( 1) ( 2)( 1) x x x x x x x x x x x x x x x x → → → − + +  − = = =  + − −  − + + + + + +   62 ( ) 2 2 0 0 (1 )(1 )(1 ) (6 11 6) lim lim lim 11 6 x x x x x x x x x x x x x → → → + + + − + + = = + + = 63 1 2 1 2 1 1 1 ( 1)( 1) lim lim lim 1 ( 1)( 1) n n n n n m m m m m x x x x x x x x x x x n x x x x x x x x m − − − − − − − − → → → − = − + + + + = + + + + = − − + + + + + + + + 64 1 2 1 1 ( 1)( 1) n( 1) lim lim ( 1) ( 1) n n n x x x nx n x x x x x x x − − → → − + − = − + + + + − − − − 1 2 1 ( 1) ( 1) ( 1) ( 1) lim 1 n n x x x x x x − − → − + − + + − + − = − ( 3 ) 1 lim ( n n 1) ( n n 1) x x x x x x x − − − − → = + + + + + + + + + + + ( 2)( 1) ( 2) ( 3) 2 n n n n − − = − + − + + + = 65 100 50 2 lim 2 x x x x x → − + − + 99 98 49 48 1 ( 1)( 1) ( 1) lim ( 1)( 1) ( 1) x x x x x x x x x x x → − + + + + − − = − + + + + − − 99 98 49 48 1 49 lim 24 x x x x x x x → + + + = = + + + 66 2 1 ( 1) ( 1) ( 1) lim lim 1 n n x x x x x n x x x x x → → + + + − − + − + + − = − − 1 1 ( 1) ( 1)( 1) ( 1)( 1) lim 1 n n x x x x x x x x x − − → − + − + + + − + + + + = − 1 1 lim(1 ( 1) ( n n 1)) x x x x x − − → = + + + + + + + + n ( 1) n n+ = + + + + = Bài Tính giới hạn sau: 1 1 3 lim 1 x x x → + − − ĐS: 1 4 2 2 lim 3 x x x →− + (53)3 3 lim x x x → − + − ĐS: 1 − 8 lim x x x → − − + ĐS: −6 5 lim x x x x →− + + − + ĐS: 1 − 2 3 lim x x x x x → − − − ĐS: 1 7 2 2 2 lim x x x → + − − ĐS: 1 16 2 2 lim x x x → − − − ĐS: 3 16 − 9 lim x x x → − + − ĐS: 24 10 9 3 lim x x x x → − − ĐS: 1 54 − 11 49 lim x x x → − − − ĐS: −56 12 1 2 lim x x x x → − + − ĐS: 7 13 2 3 lim x x x x x → − + − ĐS: 2 9 14 2 lim x x x x x → − − − ĐS:  15 2 2 4 lim x x x x → + − − ĐS: 1 3 16 3 3 lim x x x x → − − − ĐS: 1 17 2 2 2 lim 10 x x x x → + − + − ĐS: 1 4 18 2 lim 1 x x x x → − + − − ĐS: 19 4 lim x x x x → − − + − ĐS: 30 20 1 3 lim x x x x → + − + − ĐS: 1 21 2 1 lim x x x → − − ĐS: 1 4 22 2 2 3 3( 1) lim 3 x x x x → − + − + ĐS: 12− 23 1 lim x x x x → + − + ĐS: 24 2 lim x x x →− + − − ĐS: 1 − 25 2 lim x x x x x → − − − ĐS: 26 2 2 5 lim x x x x x → + + − − ĐS: 2 27 lim 2 x x x x x → − + + − ĐS: 3 4 28 2 lim x x x x →− − + − − ĐS: 1 29 2 2 lim 3 x x x x x x →− + − + + + + ĐS: 5 2 30 5 lim x x x x → − − + − ĐS: 31 1 3 lim 3 2 x x x x →− + + − + ĐS:6 32 2 2 2 6 lim 4 x x x x x x x → − + − + − − + ĐS: 1 − 33 2 (54)Page 54 35 9 3 lim 5 x x x → − − − ĐS: 2 − 36 1 3 lim 8 x x x x → + − + + − ĐS: 37 2 2 lim 1 x x x x x → + − − − − ĐS: 1 − 38 1 3 lim 4 x x x x → + − + − + ĐS: 3 39 3 1 lim 2 x x x x x → + − − + − + ĐS: −3 40 2 2 2 lim 1 x x x x x → + − + + − + ĐS: 2 41 2 1 lim 3 x x x x x → − + + − ĐS: 4 − 42 4 1 4 lim 1 x x x → + − − ĐS: 43 4 2 1 3 lim 2 x x x x x x → − + − + + − ĐS: Lời giải 1 Ta có 1 1 3 ( 2)( 2) 1 lim lim lim 1 ( 1)( 2) x x x x x x x x x x → → → + − = + − + + = = − − + + + + 2 Ta có 2 2 2 ( 2)( 1) lim lim lim ( 1) 3 ( 1)( 1) x x x x x x x x x x →− →− →− + = + + + = + + = + − + − + + 3 Ta có 6 6 3 (3 3)(3 3) 1 lim lim lim 6 ( 6)(3 3) 3 x x x x x x x x x x → → → − + = − + + + = − = − − − + + + + 4 Ta có 8 8 8 ( 8)(3 1) lim lim lim 1 3 (3 1)(3 1) x x x x x x x x x x → → → − = − + + = + + = − − − + − + + + 5 Ta có 2 2 1 1 4 ( 1) lim lim lim 1 ( 1)( 4 2) 4 2 x x x x x x x x x x x x x x →− →− →− + + − = + = = − + + + + − + + − 6 Ta có 2 2 2 3 2 ( )( ) lim lim 2 (2 6)( 2 3 ) x x x x x x x x x x x x x x x x → → − − = − − − + − − − + 2 3 ( 3) lim lim 4 2( 3)( ) 2( ) x x x x x x x x x x x x → → − = = = − − + − + 7 Ta có 2 2 2 2 2 1 lim lim lim 4 ( 2)( 2)( 2) ( 2)( 2) 16 x x x x x x x x x x x → → → + − = − = = − − + + − + + − 8 Ta có 2 2 2 2 3(2 ) 3 lim lim lim 4 ( 2)( 2)(2 2) ( 2)(2 2) 16 x x x x x x x x x x x → → → − − − = = = − − − + + − + + − 9 Ta có 2 3 3 9 ( 3)( 3)( 2) lim lim lim ( 3)( 2) 24 1 ( 2)( 2) x x x x x x x x x x x x → → → − + − + +   = =  + + + = (55)10 Ta có 2 9 9 3 1 lim lim lim 9 (9 )( 3) ( 3) 54 x x x x x x x x x x x x → → → − − − = = = − − − + + 11 Ta có 2 7 49 ( 7)( 7)(2 3) lim lim 2 (2 3)(2 3) x x x x x x x x x → → − − + + − = − − − − + − 7 ( 7)( 7)(2 3) lim lim( 7)(2 3) 56 7 x x x x x x x x → → − + + − = = − + + − = − − 12 Ta có 2 1 1 2 4 lim lim lim 1 ( 1)( 1)(2 3) ( 1)(2 3) x x x x x x x x x x x x x x x x → → → − + = − − = + = − − + + + + + + 13 Ta có 2 3 3 3 2 lim lim lim 3 ( 3)( ) ( ) x x x x x x x x x x x x x x x x x → → → − + = − − = + = − − + + + + 14 Ta có 2 2 2 1 1 ( 1)( 1) ( 1) lim lim lim 2 ( 1) 2 x x x x x x x x x x x x x x x x x → → → − = − − + = − + =  − + − − − − − 15 Ta có 2 2 2 4 4( 2) lim lim lim 2 ( 2)( 3) ( 3) x x x x x x x x x x x x → → → + − = − = = − − + + + + 16 Ta có 2 4 4 3 3 3( 4) lim lim lim 4 ( 4)( 3 3) ( 3 3) x x x x x x x x x x x x → → → − − = − = = − − − + − + 17 Ta có 2 2 2 2 ( 2)( 2) lim lim lim 2 10 ( 2)(2 5)( 2) (2 5)( 2)( 2) x x x x x x x x x x x x x x x → → → + − = + − + + = − + − − − + + − − + + 2 1 lim 4 (2 5)( 2) xx x = = − − + + 18 Ta có 2 2 2 3 ( 1)( 2)( 1) lim lim lim( 1)( 1) 1 ( 1)( 1) x x x x x x x x x x x x x → → → − + = − − − + = − − + = − − − − − + 19 Ta có 2 4 4 3 ( 1)( 4)( 3) lim lim lim( 1)( 3) 30 4 5 x x x x x x x x x x x x → → → − − = + − + + = + + + = − + − 20 Ta có 2 1 1 3 3( 1) lim lim lim 2 ( 1)( 2)( 2) ( 2)( 2) x x x x x x x x x x x x → → → + − − = = = + − − + + + + + + 21 Ta có 2 1 1 1 1 lim lim lim 1 ( 1)( 1)( 1) ( 1)( 1) x x x x x x x x x x x → → → − − = = = − + − + + + 22 Ta có 2 2 3 4( 1) ( 2)(3 2)(3 1) lim lim 3 (3 1)(3 1) x x x x x x x x x x → → − + = − + + + − + − + + + ( 2)(3 2)(3 1) lim 4(2 ) x x x x x → − + + + = − (3 2)(3 1) lim 12 4 x x x → + + + (56)Page 56 23 Ta có 3 2 3 3 0 0 1 lim lim lim ( 1)( 1) ( 1)( 1) x x x x x x x x x x x x x → → → + − = = = + + + + + + + 24 Ta có 3 2 3 2 2 2 ( 2)( 3) lim lim lim ( 2)( 4) ( 4) 1 x x x x x x x x x x x x x →− →− →− + + − + − + = = = − + − + − − + − − 25 Ta có 2 2 2 2 1 1 2 ( 1) ( 1) lim lim lim ( 1)( 1) ( 1) x x x x x x x x x x x x x x x x → → → − − = − − = − − = − − − + − + 26 Ta có 2 2 2 5 12 20 lim lim 2 ( 2)( ( 5)) x x x x x x x x x x x x → → + + − = − + − − − + − + 2 ( 2)( 10) ( 10) lim lim 3 ( 2)( ( 5)) ( ( 5)) x x x x x x x x x x x x → → − − − − − = = = − + − + + − + 27 Ta có 2 2 1 1 ( 1)( ( 4) ( ( 4) lim lim lim 10 ( 9) 2 x x x x x x x x x x x x x x x x x → → → − = − + − − = + − − = − + − − − + + − 28 Ta có 2 1 2 2 lim lim 1 ( 1)( 1)(( 2) ) x x x x x x x x x x x →− →− − + − − − = − − + − − − 1 3 lim 6 ( 1)(( 2) ) x x x x x →− + = = − − − − 29 Ta có 2 2 2 1 2 17 lim lim 3 ( 2)((2 5) 2 8) x x x x x x x x x x x x →− →− + − + + = + = + + + + + + + 30 Ta có 2 2 5 4( 2) lim lim lim 2 ( 2)( 2) ( 2) x x x x x x x x x x x x → → → − − + = − = = − − − + + − + + 31 Ta có 1 3 3( 1)( 2) lim lim 1 3 2 x x x x x x x x x →− →− + = + + + + + + − + =xlim 3( 2→−1 + x+ x+2)=6 32 Ta có 2 2 2 2 3 2 6 lim lim 4 ( 1)( 2 6 2 6) x x x x x x x x x x x x x → → − + − + − = − = − + − − + + + − 33 Ta có 2 4 2 2 1 2 1 lim lim ( 1)( ) x x x x x x x x x x x x x x →− →− + + − − + = = + − + + + + − 34 Ta có 2 2 3 lim lim 2 7 2 x x x x x x → → − + + + = − = − + − + + 35 Ta có 9 3 2 lim lim 3 5 x x x x x x → → − = − − + = − − − + 36 Ta có 1 3 2( 3) lim lim 8 3 x x x x x x x x → → + − + = + + = (57)37 Ta có 2 2 lim lim 4 1 2 x x x x x x x x x x → → + − − + − = = − − − − + + 38 Ta có 1 3 lim lim 2 4 x x x x x x x x → → + − + + + = = + − + + + 39 Ta có 3 1 lim lim 2 x x x x x x x x x x → → + − + = − + + + = − + − + + + + 40 Ta có 2 2 2 2 5 lim lim 3 1 2 x x x x x x x x x x → → + − + = + + + = + − + + + + 41 Ta có 2 3 2 1 1 ( ) lim lim 5 3 3 x x x x x x x x x x x x → → − = + − − = − − + − − + + − 42 Ta có 4 3 1 4 4 lim lim 1 (4 3) (4 3) 4 3 1 x x x x x x x → → − − = = − − + − + − + 43 Ta có 4 2 1 3 lim 2 x x x x x x → − + − + + = − Bài Tính giới hạn sau: 1 3 2 4 lim 2 x x x → − − ĐS: 1 3 3 1 5 lim 1 x x x →− − + + ĐS: 5 12 3 3 1 lim x x x x → − − + ĐS: 1 3 3 1 2 lim 1 x x x → − + − ĐS: 5 12 − 3 3 3 lim 1 x x x → − − − ĐS: 1 lim 1 x x x → − + − ĐS: 7 3 5 lim 2 x x x x x → − − − − ĐS: 2 9 1 lim 1 x x x → − − ĐS: 9 3 3 3 27 lim 1 28 x x x x → − + − + ĐS: 54 10 3 5 lim 30 x x x x → + − + − ĐS: 1 336 11 3 2 10 lim 3 x x x x x →− + + − + + ĐS: 3 2 12 3 2 1 lim 3 x x x → − + + − ĐS: 2 13 3 1 lim 7 x x x → − + − ĐS: 14 2 3 3 lim 1 x x x →− + − + ĐS: 3 − 15 3 1 2 lim 1 x x x x → − − − ĐS: 2 3 16 3 2 1 lim 2 x x x → − (58)Page 58 17 3 3 1 lim 4 x x x → − + − ĐS: 18 3 2 2 lim 1 x x x x →− + + − ĐS: 1 − 19 3 3 9 lim 1 x x x x → + + − + ĐS: 1 12 20 3 3 19 lim 4 3 x x x → − + − − ĐS: 27 − 21 3 2 1 lim 4 x x x x x → + − − − ĐS: 1 − 22 3 2 1 lim 1 x x x → − − − ĐS: 2 23 3 2 3 lim 3 2 x x x x → + − − − ĐS: 1− 24 4 1 lim 2 1 x x x → + − + − ĐS: 2 Lời giải 1) Ta có 3 3 2 4 lim lim 2 16 2 4 4 x x x x x x → → − = = − + + 2) Ta có ( ) 3 2 1 13 5 5 lim lim 1 5 3 2 5 3 4 12 x x x x x x →− →− − + = = + − − − + 3) Ta có ( )( ( ) ) 3 0 3 3 1 1 lim lim 3 1 1 x x x x x x x x → → − − = = + + + − + − 4) Ta có ( ) 3 2 1 3 3 2 5 lim lim 1 4 5 3 5 3 12 x x x x x x → → − + = − = − − + + + + 5) Ta có ( ) 2 3 2 3 3 3 1 3 lim lim 3 x x x x x x x → → − + − + − = = + − − 6) Ta có ( 3( )2) 3 1 1 lim lim 2 1 x x x x x x → → − = − − + − = + − 7) Ta có ( ) (( ) ) 2 2 1 3 3 5 4 lim lim 2 2 1 5 4 5 4 4 x x x x x x x x x x x → → − − = − − + = − − + − + − + 8) Ta có (3 ) 3 1 1 lim lim 1 x x x x x x → → − = + + = (59)9) Ta có 3 3 3 27 lim 1 28 x x x x → − + − + ( )( ) ( ) ( ) ( ) ( )( ) 2 2 3 2 3 2 3 1 28 28 lim 3 x x x x x x x x x x x →   − + +  + + + + + +    = − + + ( ) ( ) (2 )3 3( )2 2 3 1 28 28 lim 72 2 x x x x x x x x x →   + +  + + + + + +    = = + + 10) Ta có ( ) ( ) 3 3 3 5 1 lim lim 30 3 10 5 5 4 336 x x x x x x x x x → → + − = = + − + +  + + + +      11) Ta có 3 2 10 lim 3 x x x x x →− + + − + + ( )( ) ( ) ( ) ( ) 3 1 3 3 3 2 3 3 3 lim 1 10 10 x x x x x x x x x x →− − + + =   + +  + + − + + −    ( ) ( ) ( ) ( ) 2 1 3 3 3 3 3 lim 2 2 10 10 x x x x x x x x →− − + = =   +  + + − + + −    12) Ta có ( )( ) 2 2 1 3 1 2 lim lim 3 3 1 x x x x x x x x → → − = + + = + − + + + 13) Ta có ( ) 2 3 3 3 1 7 1 lim lim 7 x x x x x x x → → + + + + − = = + − + 14) Ta có ( )( ) 3 2 3 1 1 3 lim lim 2 1 x x x x x x x x →− →− − − + + − = = − + + + 15) Ta có ( ) ( ) 3 2 1 3 2 1 lim lim 3 1 2 1 2 1 x x x x x x x x x x → → − − = + = − − + − + 16) Ta có ( ) 2 3 3 3 1 2 1 lim lim 2 1 x x x x x x x x → → − − − + − = = − + + + 17) Ta có ( ) ( ) 2 3 3 3 1 3 4 4 4 1 lim lim 4 4 1 x x x x x x x x → → + + + + − = = + − + + 18) Ta có ( ) ( ) ( ) 3 2 1 3 2 lim lim 1 1 2 2 x x x x x x x x x x →− →− + + = = − − −  + − + +    (60)Page 60 19) Ta có 3 3 9 lim 1 x x x x →− + + − + ( ) ( ) ( )( ) ( ) 1 2 2 3 3 3 lim 2 1 9 6 x x x x x x x →− = =   − +  + − + − + +    20) Ta có ( )( ) ( ) 2 3 3 3 3 3 3 9 3 19 27 lim lim 8 3 4 19 19 x x x x x x x x x → → − + − + − + = = −   − − − − − − +     21) Ta có ( ) ( ) ( ) 3 2 0 3 3 2 3 1 lim lim 4 4 1 1 1 x x x x x x x x x x → → + − − = = − − −  + + − + −      22) Ta có ( ) ( ) 3 1 2 3 3 2 1 2 lim lim 1 1 2 1 2 1 1 x x x x x x x x → → − − = = − + +  − + − +      23) Ta có ( ) ( ) ( ) 2 2 3 3 2 1 2 3 lim lim 3 2 3 3 2 3 2 x x x x x x x x x x x → → − + − + + − = = −   − − + + + +     24) Ta có ( )( ) ( ) 4 4 0 3 3 2 1 1 1 lim lim 3 2 1 2 1 1 1 x x x x x x x x → → + + + + + − = =   + − + + + +     Bài Tính giới hạn sau: 1) 0 9 16 lim x x x x → + + + − ĐS: 24 2) 2 5 lim 1 x x x x → + + + − − ĐS: 4 3) 3 2 2 lim 3 x x x x → + + − − − ĐS: 5 6 4) 2 4 lim x x x x → + + + − ĐS:5 4 5) 2 2 7 lim 2 x x x x x → + + + − − ĐS: 8 3 6) 2 2 2 lim 2 x x x x x → − + − − ĐS:8 7) ( ) 6 5 84 lim 6 x x x x x → − − + − − ĐS: 74 3 8) 3 0 1 lim x x x x → + − + ĐS:0 9) 3 1 7 lim 1 x x x x → + − + − ĐS: 1 − 10) 3 2 8 11 lim 3 x x x x x → + − + − + ĐS: 7 54 11) 3 0 2 lim x x x x → + − − ĐS:13 12 12) 3 1 3 lim 1 x x x x → + − + − ĐS: 1 13) 2 1 7 lim 1 x x x x → + − − − ĐS: 7 12 14) 3 2 3 lim 2 x x x x → + − − − ĐS: 1 (61)15) 3 2 3 lim 2 x x x x → + − − − ĐS:−1 16) 3 2 2 11 lim 4 x x x x x → + + − + − ĐS: 5 72 17) 3 3 2 5 lim 1 x x x x → − − + − ĐS: 11 24 − 18) 3 2 3 24 lim 4 x x x x x → − + + − − − ĐS: 17 16 − 19) 3 2 3 lim 3 x x x x x x → − − − − − + ĐS: 5 6 20) 3 2 2 lim 1 x x x x x → − + − − − ĐS: 3 21) 3 2 1 lim x x x x x → + − − + ĐS: 1 2 22) 3 2 6 lim 2 x x x x → + − + − ĐS: 13 96 − 23) 0 1 lim x x x x → + + − ĐS:5 24) 3 0 1 lim x x x x → + + − ĐS:7 3 25) 3 1 3 2 lim 1 x x x x → + − − − ĐS: 1 12 26) 3 4 lim x x x x x → + + − + ĐS:1 27) 2 0 4 lim x x x x → + + − − ĐS: 12 − 28) 3 1 lim x x x x → + − + ĐS:1 2 29) ( ) 2 6 lim 1 x x x x x → + + − − ĐS: 11 6 30) 4 3 lim 2 x x x x x x → − + − − + − + ĐS: 5 − 31) 2 1 3 2 lim 2 x x x x x x → − − + + + − − + ĐS: 17 16 − 32) 2 2 4 lim 2 2 x x x x x x → − + + − − + − ĐS: 8 33) 3 3 1 6 2 lim 1 x x x x x x → + − − − + ĐS: 1 8 34) ( ) 3 2 2 2 lim 2 x x x x x x → − + − − + − ĐS: 1 35) 3 1 lim x x x x → + − + ĐS:1 2 36) 3 1 lim x x x x → + − + ĐS:2 Lời giải 1) 0 9 16 lim x x x I x → + + + − = Ta có 0 9 16 lim x x x I x x →  + − + −  =  +    ( )( ) ( ) ( ( )( ) ) 0 9 16 16 lim 9 16 x x x x x x x x x →  + − + + + − + +    = +  + + + +    ( ) ( ) ( ) ( ) 0 9 16 16 lim lim 9 16 16 x x x x x x x x x x x x x x → →  + − + −        = + = +  + + + +   + + + +      1 1 (62)Page 62 2) 1 2 5 lim 1 x x x I x → + + + − = − Ta có 1 2 2 lim 1 x x x I x x →  + − + −  =  +  − −   ( )( ) ( )( ) ( )( ) ( )( ) 1 2 2 2 5 lim 1 2 x x x x x x x x x →  + − + + + − + +    = +  − + + − + +    ( )( ) ( )( ) 1 2 lim 1 2 x x x x x x x →  + − + −    = +  − + + − + +    ( ) ( )( ) ( ) ( )( ) 1 2 lim 1 2 x x x x x x x →  − −    = +  − + + − + +    1 2 5 lim 4 2 2 xx x   =  +  = + = + + + +   3) 3 2 2 lim 3 x x x I x → + + − − = − Ta có 3 2 6 2 lim 3 x x x I x x →  + − − −  =  +  − −   ( )( ) ( )( ) ( )( ) ( )( ) 3 2 6 2 2 2 lim 3 3 2 x x x x x x x x x →  + − + + − − − +    = +  − + + − − +    ( ) ( )( ) ( )( ) 3 2 2 lim 3 3 2 x x x x x x x →  + − − −    = +  − + + − − +    ( ) ( )( ) ( ) ( )( ) 3 2 3 lim 3 3 2 x x x x x x x →  − −    = +  − + + − − +    3 2 2 lim 6 6 2 xx x   =  + = + = + + − +   4) 0 2 4 lim x x x I x → + + + − = Ta có 0 2 lim x x x I x x →  + − + −  =  +    ( )( ) ( ) ( ( )( ) ) 0 2 1 1 4 lim 1 x x x x x x x x x →  + − + + + − + +    = +  + + + +    ( ) ( ) ( ) 0 2 1 4 lim 1 x x x x x x x →  + − + −    = +  + + + +    2 lim 2 4 1 xx x   =  +  = + = + + + +   5) 2 2 7 lim 2 x x x x I x → + + + − = (63)Ta có ( ) 2 2 2 lim 2 x x x x x I x → − + + + − + + − = − 2 lim 2 x x x x x x →  + − + −  =  + + +  − −   ( )( ) ( )( ) ( )( ) ( )( ) 2 2 2 2 7 2 lim 2 2 x x x x x x x x x →  + − + + + − + +    = + +  − + + − + +    2 2 2 lim 4 2 xx x   = +  + = + + = + + + +   6) 2 2 2 lim 2 x x x x I x → − + − = − Ta có ( ) 2 2 2 4 lim 2 x x x x x I x → − − + − − + − = − 2 2 4 4 lim 2 x x x x x x →  − − −  =  − + +  − −   ( )( ) ( )( ) ( )( ) ( ) ( )( ) 2 4 1 1 2 2 4 1 1 2 lim lim 2 2 1 1 x x x x x x x x x x x x x → →  − − − + − +   − −      = + + = + + +  − + + −   − + +      2 4 2 lim 2 2 1 xx x   = +  + + = + + = − +   7) ( ) 6 5 84 lim 6 x x x x I x → − − + − = − Ta có ( ) 6 5 30 26 78 lim 6 x x x x x I x → − − + − − + − = − ( ) ( ) 6 26 3 5 6 lim 6 6 x x x x x x x x →  − − − − −    = + +  − − −    ( )( ) ( )( ) 6 26 3 3 lim 6 3 x x x x x x →  − − − +    = − + +  − − +    ( ) ( )( ) ( ) ( )( ) 6 26 26.2 15 lim 15 lim 6 3 3 x x x x x x x x → → − − − = + + = + + − − + − − + 6 52 52 74 15 lim 15 6 2 3 xx = + + = + + = − + 8) 3 0 1 lim x x x I x → + − + = Ta có 3 0 1 1 1 lim lim x x x x x x I x x x → →   + − + − + + − − + = =  +    ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 3 0 3 3 1 1 3 1 1 lim 1 1 1 3 1 3 x x x x x x x x x x x →  − + + + + +  + − + +   =  +  + + + + + +   (64)Page 64 ( ) ( ( () ) ) ( )2 0 3 3 1 1 2 lim lim 1 1 1 1 3 1 3 1 1 3 1 3 x x x x x x x x x x x x → →     − +  + −   −  =  +  = +  + +  + + + + + + + + + +       2 0 − = + = 9) 3 1 7 lim 1 x x x I x → + − + = − Ta có 3 1 7 2 lim 1 x x x I x → + − + − + = − 3 1 7 2 lim 1 x x x x x →  + − − +  =  +  − −   ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 3 3 3 2 2 1 3 3 3 3 7 7 2 3 2 3 lim 1 1 7 x x x x x x x x x x x →  + −  + + + +     − + + +       = +    − + + + + − + +          ( ) ( ) ( ) ( )( ) 2 1 3 3 3 3 4 7 lim 1 1 7 x x x x x x x x →    + − − +    = +    − + + + + − + +          ( ) ( ) ( )( ) 3 1 3 3 3 3 1 lim 1 1 7 x x x x x x x x →    − −    = +    −  + + + +  − + +        ( ) 2 2 1 3 3 3 3 1 lim 12 4 2 7 x x x x x x x →   + + +   =  − = − = − + + + + + +     10) 3 2 8 11 lim 3 x x x I x x → + − + = − + Ta có 3 2 2 2 8 11 3 11 3 lim lim 3 3 x x x x x x I x x x x x x → →   + − + − + + − − + = =  +  − +  − + − +  ( )( ( ) ) ( )( ( ) ) ( )( ) ( )( ) 2 3 3 2 2 2 3 3 8 11 11 11 3 7 3 7 lim 3 3 11 11 x x x x x x x x x x x x x →  + − + + + +  − + + +   =  +  − + + + − + + + + +     ( )( ( ) ) ( ) ( )( ) 2 2 3 3 9 8 11 27 lim 3 3 11 11 x x x x x x x x x x →   − +  + −  =  +  − + + + − + + + + +   (65)( ) ( )( ) (( ) ) ( )( )( ) 2 3 3 8 2 lim 1 1 11 11 x x x x x x x x x x →   −  −  =  +  − − + + − − + + + +     ( ) (( ) ) ( )( ) 2 3 3 8 lim 27 54 1 1 11 11 x x x x x x →     =  − = − = − + + − + + + +     11) 3 0 2 lim x x x I x → + − − = Ta có 3 0 2 2 2 lim lim x x x x x x I x x x → →   + − + − − + − − − = =  +    ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 3 0 3 3 2 8 2 1 1 lim 1 4 8 8 x x x x x x x x x x x →  − − + − + −  + − + +   =  +  + + + − + −     ( ) ( ) ( ( )( ) ) 0 3 3 2 1 8 lim 1 4 8 8 x x x x x x x x →   + − − −   =  +  + + + − + −     ( )2 0 3 3 2 13 lim 2 12 12 1 4 8 8 x x x x →     = + = + =  + + + − + −    12) 3 1 3 lim 1 x x x I x → + − + = − Ta có 3 1 3 2 3 2 lim lim 1 1 x x x x x x I x x x → →   + − + − + + − − + = =  +  −  − −  ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 3 3 1 2 3 2 3 3 5 2 3 2 3 lim 1 1 5 x x x x x x x x x x x →  + −  + + + +        − + + +    = +    − + + + + − + +          ( ) ( ) ( ) ( )( ) 2 1 2 3 2 3 4 3 lim 1 1 5 x x x x x x x x →    + − − +    = +    −  + + + +  − + +        ( ) ( ) ( ) ( )( ) 2 1 2 3 2 3 3 1 lim 1 1 5 x x x x x x x x →    − −    = +    − + + + + − + +        (66)Page 66 ( ) ( )2 1 2 3 2 3 3 1 1 lim 12 4 2 3 5 x x x x x →   +   =  − = − = + + + + + +     13) 2 1 7 lim 1 x x x I x → + − − = − Ta có 2 3 1 7 2 2 lim lim 1 1 x x x x x x I x x x → →   + − + − − + − − − = =  +  −  − −  ( )( ( ) ) ( ) (( ) ) ( )( ) ( )( ) 2 3 3 2 1 3 3 7 7 5 lim 1 1 7 x x x x x x x x x x x →  + − + + + + − − + −    =  +  − + − − + + + +     ( ) (( ) ) ( ) ( )( ) 2 1 3 3 4 lim 1 1 7 x x x x x x x x →   − −  + −  =  +  − + − − + + + +     ( ) (( ) ) ( )( ) 2 1 3 3 1 lim 1 1 7 x x x x x x x x →    − −  =  +  − + − − + + + +     ( )2 1 3 1 1 lim 12 12 7 x x x x x →  +    = + = + =  + + + + + −    14) 3 2 3 lim 2 x x x I x → + − − = − Ta có 3 2 3 2 3 2 lim lim 2 2 x x x x x x I x x x → →   + − + − − + − − − = =  +  −  − −  ( )( ( ) ) ( ) (( ) ) ( )( ) ( )( ) 2 3 3 2 3 3 3 2 2 2 3 2 2 3 2 lim 2 2 2 x x x x x x x x x x x →  + − + + + +  − − + −   =  +  − + − − + + + +     ( ) (( ) ) ( ) ( )( ) 2 3 3 4 3 lim 2 2 2 x x x x x x x x →   − −  + −  =  +  − + − − + + + +     ( ) ( ) (( ) ) ( )( ) 2 3 3 3 lim 2 2 2 x x x x x x x x →   −  −  =  +  − + − − + + + +     ( )2 2 3 3 3 3 lim 12 2 3 2 x x x x →   −   = − = + = −  + + + + + −  (67)15) 3 2 3 lim 2 x x x I x → + − − = − Ta có 3 2 3 2 2 lim lim 2 2 x x x x x x I x x x → →   + − + − − + − − − = =  +  −  − −  ( )( ( ) ) ( ) (( ) ) ( )( ) ( )( ) 2 3 3 2 3 3 3 2 2 2 5 6 2 5 6 lim 2 2 2 x x x x x x x x x x x →  + − + + + +  − − + −   =  +  − + − − + + + +     ( ) (( ) ) ( ) ( )( ) 2 3 3 4 3 lim 2 2 2 x x x x x x x x →   − −  + −  =  +  − + − − + + + +     ( ) ( ) (( ) ) ( )( ) 2 3 3 3 10 lim 2 2 2 x x x x x x x x →   −  −  =  +  − + − − + + + +     ( )2 2 3 3 3 5 lim 12 2 3 2 x x x x →   −   = − = + = −  + + + + + −    16) 3 2 2 11 lim 4 x x x x I x → + + − + = − Ta có 3 2 2 2 2 11 3 11 3 lim lim 4 4 x x x x x x x x I x x x → →   + + − + − + + + − − + = =  +  −  − −  ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 3 3 2 2 2 2 3 2 3 2 11 11 11 3 7 3 7 lim 4 4 11 11 x x x x x x x x x x x x x x x x →  + + −  + + + + + +        − + + +    = +    − + + + + + + − + +          ( ) ( ) ( ) ( )( ) 2 2 2 2 2 3 2 3 9 2 11 27 lim 4 4 11 11 x x x x x x x x x x x →    + + − − +    = +    − + + + + + + − + +          ( ) ( ) ( )( ) 2 2 2 2 2 3 2 3 2 16 lim 4 4 11 11 x x x x x x x x x x x →    + − −    = +    −  + + + + + +  − + +        ( )( ) ( ) ( ) ( )( ) 2 2 lim 4 x x x x x x →    − + −    = +   (68)Page 68 ( ) ( ) ( ) ( )( ) 2 2 3 2 3 2 12 lim 108 24 72 2 2 11 11 x x x x x x x x x →    +  −   = − = + =    + + + + + + + + + +          17) 3 3 2 5 lim 1 x x x I x → − − + = − Ta có 3 3 2 2 1 5 2 2 lim lim 1 1 x x x x x x I x x x → →   − − + − + − − − + = =  +  −  − −  ( )( ) ( )( ) ( ) ( ) ( ) ( ) 2 3 3 3 1 2 3 2 2 3 2 7 5 lim 1 1 4 2 7 7 x x x x x x x x x x x →  − +  + + + +    − − − +      = +    − − + −  + + + +         ( )( ) ( ) ( ) ( ) 2 1 2 3 2 2 3 8 5 lim 1 1 4 2 7 7 x x x x x x x x →    − − − +    = +    − − + − + + + +         ( )( ) ( ) ( ) 3 1 2 3 2 2 3 1 lim 1 1 4 2 7 7 x x x x x x x x →    − −    = +    − − + −  + + + +       ( ) ( )( ) ( ) 2 2 1 3 2 2 3 1 1 3 1 11 lim 8 12 24 1 4 2 7 7 x x x x x x x →  − + +  − −   =  + = − + = − + − + + + + +     18) 3 2 3 24 lim 4 x x x x I x → − + + − − = − Ta có 3 2 3 24 2 8 lim 4 x x x x I x → − − + + − + − − = − 1 3 2 2 2 2 3 24 2 8 lim lim lim 4 4 x x x I I I x x x x x x → → → − − + − − − = + + − − − 1 I 3 2 3 24 lim 4 x x x → − − = − ( ) ( ) 2 3 3 3 2 2 2 3 3 3 3 2 3 24 24 24.2 lim 4 24 24.2 x x x x x x x →   − −  − + − +    =   −  − + − +  (69)( ) ( ) ( ) 3 2 3 2 3 3 24 lim 4 24 24.2 x x x x x → − − =   −  − + − +    ( ) ( ) ( ) 3 2 3 2 3 3.4 lim 4 24 24.2 x x x x x → − =   −  − + − +    ( )( ) ( )( ) ( ) 2 2 3 3 12 2 lim 2 24 24.2 x x x x x x x x → − + + =   − +  − + − +    ( ) ( ) ( ) 2 2 2 3 3 3 3 2 12 lim 2 24 24.2 x x x x x x → − + + =   +  − + − +    144 48 = − = − 2 I 2 2 2 lim x x x → + − = − ( )( ) ( 2)( ) 2 2 2 lim 4 2 x x x x x → + − + + = − + + ( ) ( )( )( ) 2 2 lim 2 2 x x x x x → + − = − + + + ( )( ) 1 lim 16 2 2 x→ x x − = = − + + + 3 I 2 2 8 lim x x x → − − = − ( )( ) ( 2)( ) 2 8 3 lim 4 x x x x x → − − + − = − + − ( ) ( ) ( )( ) 2 8 lim 2 x x x x → − − = + − ( ) ( )( )( ) 2 8.2 lim 2 2 x x x x x → − = − + + + 2( )( ) 16 lim 2 1 x x x → = + + + − 16 = = 1 3 16 I = − − + 17 16 = − Bài Tính giới hạn sau: 1 ( ) lim x→+ xx ĐS: + ( ) 3 lim x→− xx + ĐS: − 3 ( ) lim x→+ − −x x + x+ ĐS: − ( ) 3 lim x→− − +x x− ĐS: + 5 ( ) lim x→+ xx + ĐS: + ( ) 4 lim 10 x→− xx + ĐS: + 7 ( ) lim x→+ − +x x + ĐS: − ( ) 4 lim x→− − − +x x ĐS: − 9 lim x→ xx+ ĐS: + 10 ( ) 2 lim x→− x + +x ĐS: + 11 ( ) lim x→− x + + +x x ĐS: − 12 ( ) 2 lim x→+ x + + −x x ĐS: + (70)Page 70 14 lim( 16 3) x→− x+ + x+ ĐS: không tồn giới hạn Lời giải 1 lim 2( 3 ) x I x x →+ = − Ta có lim 2( 3 ) x I x x →+ = − 2 3 lim x→+x x   =  − = +   (vì 3 lim x→+x = + 3 lim 2 x→+ x  − =      ) lim( 3 2) x I x x →− = − + Ta có lim( 3 2) x I x x →− = − + 3 3 lim x→−x x x   =  − + = −   (vì 3 lim x→−x = − 3 3 lim 1 x→− x x  − + =     ) 3 lim( 1) x I x x x →+ = − − + + Ta có lim 92 13 x I x x x x →+   = − − + + = −   (vì lim x→+x = + 6 lim 1 x→+ x x x − − + + = −      ) 4 lim( 3 1) x I x x →− = − + − Ta có lim 32 13 x I x x x →−   = − + − = +   (vì 3 lim x→−x = − 3 lim 1 x→− x x − + − = −      ) 5 lim( 2 1) x I x x →+ = − + Ta có lim 22 14 x I x x x →+   =  − + = +   (vì 4 lim x→+x = + 2 lim 1 x→+ x x  − + =      ) 6 lim( 10) x I x x →− = − + Ta có lim 82 104 x I x x x →−   =  − + =    (vì 4 lim x→−x = + 8 10 lim 1 x→− x x  − + =      ) 7 lim( 2 3) x I x x →+ = − + + Ta có 2 2 lim x I x x x →+   = − + + = −   ( 4 lim x→+x = + 2 lim 1 x→+ x x − + + = −      ) 8 lim( 6) x I x x →− = − − + Ta có 2 1 lim x I x x x →−   = − − + = −   (vì 4 lim x→−x = + 1 lim 1 x→− x x − − + = −     ) 9 lim x I x x → = − + Ta có lim 42 x I x x x →   =  − +    3 lim x→ x x x   =  − + = +   (vì lim x→ x = + 3 lim 1 x→ x x  − +  =      ) 10 ( ) lim x I x x →− (71)Ta có lim( 2 ) x I x x →− = + + lim 12 x→−x x   = − + + = +   (vì lim x→−x= − 1 lim 2 x→− x   − + + = − +        ) 11 ( ) lim x I x x x →− = + + + lim 1 12 x→−x x x   = − + + + = −   (vì lim x→−x= −, 1 lim x→− x x   − + + + =        ) 12 lim( ) x I x x x →+ = + + − lim 12 x→+x x x   =  + + − = +   (vì lim x→+x= +, 1 lim 1 x→+ x x   =  + + − =    ) 13 lim( 1) x I x x →+ = + − + lim 1 x→+ x x x   =  + − + = −   (vì lim x→+ x = +, 1 lim x→+ x x x   =  + − + = −   ) 14 lim ( 16 3) x I x x →− = + + + Tập xác định hàm số f x( )= 16x+ +7 9x+3 1; D= − +     Ta có x→ − hàm số f x( )= 16x+ +7 9x+3 không xác định Do ( ) lim 16 x→− x+ + x+ không tồn Bài Tính giới hạn sau: 1 lim x x x →+ + − ĐS: 2 lim 1 x x x →− + ĐS: 3 lim x x x →+ − − ĐS: 1 − lim 1 x x x →− − + ĐS: 5 3 3 2 lim 1 x x x x x →+ + − − − + ĐS: −2 ( ) ( )( ) 2 2 3 lim 5 x x x x x x →+ − − + ĐS: 6 7 4 4 2 15 lim 1 x x x x →− + − + ĐS: ( )( ) ( )( ) 2 3 4 lim 2 x x x x x →+ + − − + ĐS: 9 ( ) ( ) ( ) 2 4 1 lim 3 x x x x →− − + + ĐS: 25 81 10 ( ) ( ) ( ) ( ) 4 5 1 lim 2 x x x x x →− + − + + ĐS: 1 − 11 ( ) ( ) ( )( ) 2 2 2 lim 2 1 x x x x x →− + + + − ĐS: − 12 ( ) ( ) ( ) 3 5 2 lim 1 x x x x x →− + − − ĐS: 1 32 (72)Page 72 13 2 lim 3 x x x x x →−   −  − +    ĐS: 2 14 3 lim x x x x →− − + − ĐS: 15 2 lim x x x x x →+ + + + + ĐS: 16 ( )( ) ( )( ) 2 3 4 lim 2 x x x x x →+ + − − + ĐS: 17 ( )( ) 2 2 4 lim x x x x x →− + + − + ĐS: − 18 3 2 lim x x x x x →+ + + + + ĐS: + 19 3 2 lim x x x x x x →− + + + + + ĐS: − 20 4 3 2 lim x x x x x x →+ + + + − ĐS: − 21 11 lim x x x x →+ − + − ĐS: + 22 4 2 lim x x x x →+ + − − ĐS: + 23 lim x x x x →+ − − ĐS: + 24 ( )( ) 5 3 lim x x x x x x →+ + − − + ĐS: 25 3 lim x x x x →+ + + + ĐS: 26 4 2 lim x x x x →+ + − − ĐS: − Lời giải 1 lim x x I x →+ + = − lim 1 x x x x x →+  +      =  −      lim 1 x x x →+  +      =  −      = 2 lim x x I x →− = + lim 1 x x x x →−   +     lim 1 x x →− = + = 3 lim x x I x →+ − = − 1 lim x x x x x →+  −      =  −      1 lim 2 x x x →+ − = = − − 4 lim x x I x →− − = + lim 1 x x x x x →−  −      =  +      lim 1 x x x →−  −      =  +      = 3 2 lim x x x I x x →+ + − = − − + 3 3 lim 1 x x x x x x x →+  + −      = − − +      3 lim 1 x x x x x →+  + −      = − − − +      2 lim x x x x I x x x x →+  −      =  −   +          lim 1 (73)7 4 4 2 15 lim x x x I x →− + − = + 4 4 15 lim 1 x x x x x x →−  + −      =  +      4 15 lim 1 x x x x →−  + −      = =  +      8 ( )( ) ( )( ) 2 3 4 lim 2 x x x I x x →+ + − = − + 2 3 1 lim x x x x x x x x x →+  +   −          =  −   +          1 lim x x x x x x x →+  +   −          = =  −   +          9 ( ) ( ) ( ) 2 4 1 lim x x x I x →− − + = + 2 2 4 lim x x x x x x x →−  −   +          =  +      2 25 lim 81 x x x x →−  −   +          = =  +      10 ( ) ( ) ( ) ( ) 4 5 2 1 lim 2 x x x I x x →− + − = + + 4 5 2 1 lim x x x x x x x x x →−  +   −          =  +   +          1 lim x x x x x →−  +   −          = = −  +   +          11 ( ) ( ) ( )( ) 2 2 2 lim 2 1 x x x I x x →− + + = + − 2 2 2 1 lim 1 x x x x x x x x x →−  +   +          =  +   −          2 2 2 1 lim 1 x x x x x x →−  +   +          = = −  +  −        (vì lim x→−x= −, 2 2 2 1 lim 1 x x x x x →−  +   +          =   +  −        ) 12 ( ) ( ) ( ) 2 lim x x x I x x →− + − = − 4 5 2 1 lim x x x x x x x x →−  +   −          =  −      1 1 lim 32 x x x x →−  +   −          = = −  −      13 2 lim 3 x x x I x x →−   =  −  − +   Ta có 2 lim 3 x x x I x x →−   =  −  − +   ( ) ( ) ( )( ) 3 2 2 3 lim 3 x x x x x x x →− + − − = − + ( )( ) 2 lim 3 x x x x x →− + = − + lim 3 x x x x x x x →−  +      =  −   +          2 lim 4 3 x x x x →−  +      = =  −  +        14 lim x x I = − + − 2 2 3 lim x x x x  − +      =   3 lim x x x  − +      = (74)Page 74 Bài Tính giới hạn sau: 1 lim x x x + → − − ĐS: − 2 15 lim x x x + → − − ĐS: − 3 lim x x x − → − − ĐS: + ( )2 4 lim x x x − → − − ĐS: − 5 lim x x x − → − + − ĐS: + 3 lim x x x − → − − ĐS: − 7 lim x x x + → − − ĐS: − 1 lim x x x + → + − ĐS: + 9 3 lim 15 x x x + → − − ĐS: 1 5 10 ( )3 7 lim x x x − → − − + ĐS: − 11 2 2 2 lim 2 x x x x − → − − + ĐS: 1 3 12 1 lim x x x x + → − + − ĐS: 1 13 3 1 lim x x x x − → − + − ĐS: 1 − 14 2 lim x x x x + → − + − ĐS: 15 lim x x x → − − ĐS: không tồn 16 4 4 lim 20 x x x x → − + − ĐS: không tồn 17 2 lim 1 x x x − → − − − ĐS: 18 3 lim 5 11 x x x − → − − − ĐS: 4 − 19 2 lim 1 x x x − → − − − ĐS: −3 20 2 25 lim x x x − → − − − ĐS: −30 21 ( )2 3 lim x x x + → − − ĐS: + 22 3 2 25 lim x x x x → + − − − ĐS: 1 81 23 2 lim 16 x x x + → + − ĐS: + 24 lim x x x x x + → + − ĐS: −1 25 2 lim x x x + → − − ĐS: 26 2 lim x x x x x + → + − ĐS: −2 27 ( ) ( )( ) lim 1 x x x x x + → − + + + − − ĐS: 28 2 lim x x x x − → − + − ĐS: 1 − 29 2 lim x x x x x − → − + − + − ĐS: − 30 ( ) 2 lim x x x x x + → − + + + ĐS: 31 ( ) 2 2 lim x x x x + → − − ĐS: 32 ( ) ( ) 3 lim 1 x x x x + → − + − ĐS: 33 ( ) 2 1 lim x x x x x + → + − + − ĐS: 34 1 lim 2 1 x x x x x − → − − + − ĐS: 1 35 lim x x x x + →  −        ĐS: 36 ( ) ( ) 2 2 lim x x x x + → − + − (75)37 2 2 1 lim 2 x→ − x x  −   − −    ĐS: − 38 3 lim x x x x x − → − + − + ĐS: 3 − Lời giải 1 lim x x x + → − = − − ( ) ( ) 1 lim lim 1 0, x x x x x x + + → → + − = −   − =   −   →  2 15 lim x x x + → − = − − ( ) ( ) 2 2 lim 15 13 lim 2 0, x x x x x x + + → → + − = −   − =   −   →  lim x x x − → − = + − ( ) ( ) 3 3 lim lim 3 0, x x x x x x − − → → − − = −   − =   −   →  ( )2 lim x x x − → − = − − ( ) ( ) ( ) 4 lim lim 4 0, x x x x x x − − → → − − = −    − =    −   →  lim x x x − → − + = + − ( ) ( ) 2 2 lim lim 2 0, x x x x x x − − → → − − + = −   − =   −   →  lim x x x − → − = − − ( ) ( ) 1 1 lim lim 1 0, x x x x x x − − → → − − =   − =   −   →  lim x x x + → − = − − ( ) ( ) 2 2 lim lim 4 0, x x x x x x + + → → + − = −   − =   −   →  lim x x x + → + = + − ( ) ( ) 2 2 lim lim 2 0, x x x x x x + + → → + + =   − =   −   →  9 Do x→3+ nên x− = −3 x suy 3 3 lim lim 5 15 15 x x x x x x + + → → − = − = − − 1 lim 5 x→+ = 10 ( )3 7 lim x x x − → − − = − + ( ) ( ) ( ) 3 3 lim 22 lim (76)Page 76 11 Do x→2− nên 2− = −x x suy ( )( ) 2 2 2 lim lim 2 2 x x x x x x x x − − → → − = − − + − − 2 1 lim 2 x→− x = = − 12 Do x→1+ nên x− = −1 x suy ( )( ) 3 1 1 lim lim 2 2 x x x x x x x x x + + → → − = − + − − + + 2 1 lim 2 x→+ x x = + + 13 Do x→1− nên x− = −1 x suy ( )( ) 3 1 1 lim lim 2 2 x x x x x x x x x − − → → − = − + − − + + 2 1 lim 2 x→− x x − = − + + 14 Ta có ( )( ) 3 2 xx+ = xx− , x→2+ nên 3 xx+  , suy ( )( ) ( ) 2 2 2 3 2 lim lim lim 1 2 x x x x x x x x x x + + + → → → − + − − = = − = − − 15 Ta có ( )( ) 2 3 9 3 lim lim 3 x x x x x x x → → − + − = − − TH1: x3 ta có ( )( ) ( ) 2 3 3 9 3 lim lim lim 3 x x x x x x x x x + + + → → → − + − = = + = − − TH2: x3 ta có ( )( ) ( ) 2 3 3 9 3 lim lim lim 3 x x x x x x x x x − − − → → → − − + − = = − − = − − − Do 2 3 9 lim lim 3 x x x x x x + − → → − −  − − nên không tồn 3 9 lim 3 x x x → − − 16 Ta có ( )( ) 2 4 4 lim lim 4 20 x x x x x x x x → → − = − − + + − TH1: x4, ta có ( )( ) 2 4 4 4 1 lim lim lim 4 5 20 x x x x x x x x x x + + + → → → − = − = = − + + + − TH2: x4, ta có ( )( ) 2 4 4 4 1 lim lim lim 4 5 20 x x x x x x x x x x + − − → → → − = − = − =− − + + + − Do 2 4 lim 20 x x x x + → − + −  4 4 lim 20 x x x x − → − + − nên không tồn 4 4 lim 20 x x x x → − + − 17 Do x→2− nên x− = −2 x suy 2 2 lim 1 x x x − → − − − ( )( ) 2 2 1 lim 1 x x x x − → − − + = − − ( ) 2 lim 1 x→ − x− + = 18 Do x→3− nên x− = −2 x suy 3 3 lim 5 11 x x x − → − − − ( )( ) 3 3 11 lim 5 11 x x x x − → − − + = − − ( ) 3 5 11 4 lim 5 x x − → − − + (77)19 Do x→2− nên x− = −2 x suy 2 lim 1 x x x − → − − − ( )( ) 3 2 2 1 lim 1 x x x x x − → − − + − + = − − ( ) ( ) 3 2 lim 1 x→− x x = − − + − + = − 20 Ta có ( )( ) 25 5 x − = xx+ , x→5− nên x2−250, suy 2 25 lim x x x − → − − − ( 2)( ) 3 5 25 4 lim 4 x x x x x − → − − + − + = − − ( ( ))( ) 3 lim 4 30 x x x x − → = − + − + − + = − 21 ( )2 3 lim x x x + → − = + − , ( ) ( ) ( ) 3 lim lim 3 0, x x x x x x + + → → + − =    − =    −   →  22 Ta có 3 2 25 lim x x x x → + − − − 2( )( )( ) 3 25 27 lim 2 25 25 x x x x x x → + − = − + + + + + ( )( ) 2 3 1 lim 81 1 25 25 x x x x → = = + + + + + 23 2 lim 16 x x x + → + = + − , ( ) 2 2 2 lim lim 16 4 16 0, x x x x x x + + → → + + =    − =    −   →  24 lim x x x x x + → + − ( ) ( ) 0 1 lim lim 1 x x x x x x x x + + → → + + = = = − − − 25 2 lim x x x + → − − ( )( ) ( ) 2 2 lim lim 2 2 x x x x x x x + + → → − + = = − + = − 26 lim x x x x x + → + − = ( ) ( ) 0 2 lim lim 1 x x x x x x x x + + → → + + = = − − − 27 Ta có ( ) ( )( ) lim 1 x x x x x + → − + + + − − ( )1 ( ) ( )1 2 lim lim 1 1 1 x x x x x x x x + + → − → − + + + = = = − + + − + 28 Ta có x2−6x+ =9 (x−3)2 = −x 3, x→3− nên x2−6x+ = −9 x, suy 2 lim x x x x − → − + − ( )( ) 2 3 3 6 1 lim lim lim 9 3 x x x x x x x x x x − − − → → → − + − − = = = = − − − + + 29 Do x→1− nên x− 1 0, từ ta có 2 lim x x x x x − → − + − + − ( )( ) ( )( ) 1 lim x x x x x − → − − = − − − ( )( ) 1 lim x x x x x − → − − = − − − ( ) 3 lim x x x x − → − = − − lim −x (78)Page 78 lim x x x − → − = − − 1 lim 1 x→− x   = +  −    30 ( ) lim x x x x x + → − + + + ( ) ( )( ) 1 lim x x x x x + → − + + = + ( ) ( ) 1 lim x x x x + → − + + = = 31 ( ) 2 2 lim x x x x + → − − ( ) ( )( ) ( ) 2 lim lim 2 2 x x x x x x x x x + + → → − = − = = − + + 32 Ta có ( ) ( ) lim 1 x x x x + → − + − ( ) ( )( ) ( )( ) lim 1 1 x x x x x x x + → − = + − + − + ( ) ( ) ( ) 1 lim 1 x x x x x x + → − + = − + = − 33 Do x→1+ nên 1− x 0, ta có ( ) 2 1 lim x x x x x + → + − + − ( )( ) ( )( ) lim x x x x x + →  + −    =  − +    ( )( ) lim x x x x + →  + −    = =  +    34 ( ) 1 1 lim lim 2 1 x x x x x x x x x x − − → → − = − − + − − + − 1 lim 2 x x x − → = = + − 35 lim x x x x + →  −        ( ) lim x x x x + →  −    =     ( ) lim x x x + → = − = 36 ( ) ( ) 2 2 lim x x x x + → − + − + ( ) ( )( ) ( )2 ( ) 3 2 lim lim 3 x x x x x x x + + → − → − − + − = = = − + + 37 2 2 1 lim 2 x→ − x x  −   − −    ( )( ) 2 lim 2 x x x x − →  + −  =   − +   1 lim 2 x x x x − → +   =  = − + −   38 Do x→1− nên x− 1 0, suy (x−1)2 = − = −x 1 x nên ta có 3 lim x x x x x − → − + − + ( )( ) ( )( ) 2 lim x x x x x − → + − = − + ( ) ( )( ) 1 lim x x x x x − → − + = − + 2 lim x x x − → − + = = − + Bài Tính giới hạn sau: 1) sin lim x x x → ĐS: 2) tan lim x x x → ĐS: 2 3) 2 0 cos lim x x x → − ĐS: 2 4) sin sin sin lim 45 x x x x x → ĐS: 1 5) 0 1 cos lim 1 cos x x x → − − 6) 1 cos lim sin x x x x → − ĐS: 7) ( ) 0 sin lim cos x x ax a ax → −  ĐS: 2 a 8) 1 cos lim cos x ax bx → − − ĐS: 2 a b 9) 2 ( ) 0 1 cos lim ; x x a x → −  ĐS: 2 2 a 10) 3 (79)11) 3 0 tan sin lim sin x x x x → − ĐS:1 2 12) sin sin lim x a x a x a → − − ĐS: cosa 13) limcos cos x b x b x b → − − ĐS:−sinb 14) 1 lim sin x x x → − + ĐS: 2 − 15) ( ) ( ) 0 cos cos lim x a x a x x → + − − ĐS: −2sina 16) limtan tan x c x c x c → − − ĐS: 1 cos c 17) 3 0 1 cos lim sin x x x x → − ĐS: 2 18) 2 2 sin sin lim x a x a x a → − − ĐS: sin 2 a a 19) 2 0 cos cos lim x x x x   → − ĐS: 2 2  − 20) ( ) 3 2 8 lim tan x x x →− + + ĐS:12 21) 0 1 cos cos cos lim 1 cos x x x x x → − − ĐS:1422) ( ) ( ) 2 sin 2sin sin lim x a x a x a x → + − + + ĐS:−sin( ) 23) 0 sin tan lim ;( 0) ( ) x ax bx a b a b x → + +  + ĐS: 24) 0 cos cos cos lim x x x x x → − ĐS: 33 2 − 25) 0 cos cos cos lim 1 cos x ax bx cx x → − − ĐS: 2 2 2 bac 26) ( ) ( ) 0 sin sin lim tan( ) tan( ) x a x a x a x a x → + − − + − − ĐS: 3 cos a 27) 3 0 2 1 lim sin x x x x → + − + ĐS: 28) 2 sin sin sin lim x x x x x → − ĐS: 29) 2 cos lim 2 x x x   →− + ĐS: 30) 0 2 sin sin lim 1 sin x x x x x → −  −      ĐS: -1 31) 2 1 cos lim x x x x → + − ĐS: 32) 3 0 1 tan sin lim x x x x → + − + ĐS: 33) 2 0 1 cos cos lim sin 11 x x x x → − ĐS: 37 121 34) 3 lim tan( 1) x x x → + − − ĐS: 1 35) ( )2 1 cos lim x x x   → + − ĐS: 1 2 36) sin( 1) lim 4 x x x x → − − + ĐS: 1 − 37) 2 1 cos lim x x x x → + − ĐS: 2 38) 1 cos cos lim x x x x → − ĐS:3 Lời giải. 1) 0 sin sin lim lim 5 5 x x x x x x → →   =   =   (80)Page 80 3) 2 2 0 0 2 sin sin 1 cos 2 lim lim lim 4 2 x x x x x x x x x → → →                − =   =   =                   4) 3 0 sin sin sin sin sin sin lim lim 45 3 x x x x x x x x x x x x → →   =     =   5) 2 2 0 2 5 2sin sin 1 cos 2 25 2 2 25 lim lim lim 3 1 cos 9 2sin sin 2 2 x x x x x x x x x x x → → →             − = =     =     −               6) 2 2 2 0 0 1 cos sin 4sin cos sin lim lim lim lim cos sin sin x x x x x x x x x x x x x x x x → → → → − = = =   =     7) 2 2 0 2 sin sin sin 2 lim lim lim 1 cos 2sin sin 2 x x x ax x ax x ax ax a ax ax ax ax a a → → →           = =      = −           8) 2 2 2 2 0 2 2sin sin 1 cos 2 2 2 lim lim lim 1 cos 2sin sin 2 2 x x x ax ax bx ax a a bx ax bx bx b b → → →             − = =     =     −               9) 2 2 2 0 0 2sin sin 1 cos 2 2 lim lim lim 4 2 x x x ax ax ax a a ax x x → → →         − = =    =             10) Ta có 3 (3 ) 0 sin cos sin tan lim lim cos x x x x x x x x x → → − − = 2 3 0 2sin sin sin 2 sin 1 2 lim lim cos cos 2 x x x x x x x x x x x → →     −       = = −     = −           11) ( ) ( ) ( ) 3 0 0 sin cos tan sin 1 lim lim lim sin cos sin cos cos cos x x x x x x x x x x x x x → → →   − − = = =    +  −   12) Ta có 2 cos sin sin sin 2 2 lim lim x a x a x a x a x a x ax a → + − − = − − sin lim cos cos 2 2 x a x a x a a x a → −    +  =   − =   (81)13) Ta có 2 sin sin cos cos 2 2 lim lim x b x b x b x b x b x b x b → → + − − − = − − sin lim sin sin 2 2 x b x b x b b x b → −    +  = −  − = −     14) Ta có ( ) 0 1 1 lim lim sin sin 2 1 2 1 x x x x x x x → → − + = − − + + 1 lim sin 2 1 x x x x →   = −  = − + +   15) Ta có 0 2 sin sin cos( ) cos( ) 2 2 lim lim x x a x a x a x a x a x a x x x → → + + − + − + − + − − = 0 sin lim 2sin 2sin x x a a x →   = −  = −   16) ( ) ( ) ( ) sin sin tan tan 1 lim lim lim cos cos cos cos cos x c x c x c x c x c x c x c x c x c x c x c c → → → −  −  − = =   = − −  −  17) Ta có ( )( ) 2 0 1 cos cos cos 1 cos lim lim sin sin x x x x x x x x x x → → − + + − = ( ) 2 2 0 2sin cos cos sin 1 cos cos 2 lim lim 2 2 sin cos cos 2 2 x x x x x x x x x x x x x → →   + +  + +  = =    =     18) Ta có ( )( ) 2 2 1 cos cos sin sin 2 2 lim lim x a x a x a x a x a x a x a → → − − − − = − − + ( ) ( ) 2sin sin cos cos lim lim 2( )( ) 2( )( ) x a x a a x a x a x x a x a x a x a → → − + − − = = − + − + sin( ) sin( ) sin lim 2 x a a x a x a x a a x a → + −   =   = + −   19) Ta có ( ) ( ) 2 0 2sin sin cos cos 2 2 lim lim x x x x x x x x       → → + − − − = ( ) ( ) ( ) ( ) 2 0 sin sin 2 lim 2 2 2 x x x x x               → + −    + −  − = −     = + −       20) Ta có ( ) ( )( ) ( ) ( ) 2 2 2 2 2 2 8 lim lim lim 12 tan tan( 2) tan( 2) x x x x x x x x x x x x x →− →− →− + − +  +  + = =  − +  = (82)Page 82 2 0 2 2 3 2sin 2sin 2 lim cos cos cos 2sin 2sin 2 x x x x x x x x →     =  + +      2 2 2 0 3 sin sin 2 2 2 lim cos cos cos 14 3 sin sin sin 2 2 x x x x x x x x x x x x →                     = +     +     = + + =                     22) Ta có ( ) 2 ( ) 2 ( ) 0 sin 2sin( ) sin 2sin cos 2sin lim lim x x a x a x a a x x a x x x → → + − + + + − + = ( ) 2 0 4 sin sin 2 sin( )(cos 1) 2 lim lim x x x a x a x x x x → → − + + − = = ( ) 2 0 sin 1 2 lim sin sin 4 x x a x a x →           = − +    = −           23) Ta có 0 0 sin tan sin tan sin tan lim lim lim ( ) ( ) x x x ax bx ax bx ax bx a b ax bx ax bx ax bx a b a b x a b x a b a b → → →  +    + + = = = = + + + + 24) Ta có 2 2 0 cos cos cos cos cos cos cos cos lim lim x x x x x x x x x x x x → → − = − + − 2 2 0 7 sin sin cos sin 2 sin sin cos (1 cos ) 2 lim lim x x x x x x x x x x x x → → − + − = = 2 0 7 sin sin sin 49 2 49 33 lim cos 7 4 2 2 x x x x x x x x →           =   −     = − = −           25) Ta có 2 2 0 cos cos cos cos cos cos cos cos lim lim x x ax bx cx ax bx bx bx cx x x → → − − + − = ( ) ( ) 2 0 ( ) ( ) 2sin sin cos (1 cos ) 2sin sin cos sin 2 2 2 lim lim x x b a x b a x a b x a b x cx bx cx bx x x → → − − + + + − − = = ( ) ( ) 2 2 2 2 2 2 0 ( ) sin sin sin 2 2 lim 2 cos ( ) 4 2 2 2 x b a x a b x cx b a c b a c b a c bx a b x b a x cx →  + −     −    − − −   =   +  −     = − = −         (83)26) Ta có ( ) ( ) ( ) ( ) 0 sin( ) sin( ) cos sin lim lim sin tan tan cos cos x x a x a x a x x a x a x a x a x → → + − − = + − − + − ( ) ( ) 0 cos cos cos lim cos cos x a a x a x a x → + − = = 27) Ta có 3 0 2 1 1 1 lim lim cos sin x x x x x x x x → → + − + = + − + − + ( ) 2 2 3 0 2 2 1 1 1 1 lim sin x x x x x x x → − + + + + + + + = ( )2 3 0 2 2 1 1 1 1 lim sin x x x x x x x x → − + + + + + + = = 28) Ta có ( ) 2 4 0 sin sin 2sin cos sin sin sin lim lim x x x x x x x x x x x → → − −  = ( ) 4 0 3 sin sin sin sin 2 sin sin cos cos 2 2 lim lim x x x x x x x x x x x x → → − = = 0 3 sin sin sin sin 2 2 lim 3 2 2 x x x x x x x x x →     =       =     29) 2 sin cos lim lim 2 x x x x x x      →− →−  +      = = + + 30) ( ) 0 2 0 sin cos sin sin sin cos lim lim lim cos cos 1 2sin x x x x x x x x x x x x x x x → → → − − = =   − = −    −        31) Ta có 2 2 0 1 cos 1 cos lim lim x x x x x x x x → → + − = + − + − 2 0 1 1 cos lim x x x x x →  + − −  =  +    ( ) 2 2 0 2 2sin 1 2 lim 1 x x x x x x →    + −  =  +  + +     2 2 sin 1 2 1 lim 2 1 2 2 x x x x →           = +   = + =  + +          (84)Page 84 32) ( ) 3 3 0 1 tan sin tan sin lim lim 1 tan sin x x x x x x x x x x → → + − + + − − = + + + ( ) ( ) 3 sin cos lim cos tan sin x x x x x x x → − = + + + ( ) 2 3 2sin sin lim cos tan sin x x x x x x x → = + + + ( ) 2 0 sin 2 sin 2 lim 4 cos tan sin 2 x x x x x x x x →           =     =  + + +          33) 2 2 ( ) 0 1 cos cos cos cos cos lim lim sin 11 sin 11 x x x x x x x x x → → − + − − = 2 2 0 5 2sin cos sin 2 lim sin 11 sin 11 x x x x x x →     =  +      2 2 0 5 sin sin 25 2 11 49 2 11 lim cos 5 242 sin11 242 sin11 2 x x x x x x x x x x →                   =     +                       25 49 37 242 242 121 = + = 34) ( ) ( )( ) 1 3 lim lim tan tan 1 3 2 x x x x x x x → → + − = + − − − + + ( ) 1 1 lim tan 3 x x x x →  −  =   = − + +   35) ( ) ( ) 2 2 2 cos 1 cos 2 lim lim x x x x x x     → → + = − − ( ) 2 2 2sin sin 1 2 lim lim 2 2 x x x x x x       → →   −  −                    = =   = −   −           36) ( ) ( ) ( )( ) ( ) 2 1 1 sin sin sin 1 lim lim lim 4 3 x x x x x x x x x x x x → → → − −  −  = =  = − − + − −  − −  37) 2 2 0 1 cos 1 cos lim lim x x x x x x x x → → + − + − + − = ( ) 2 2 2 2 0 2 1 1 cos 1 2sin lim lim 1 x x x x x x x x x x x → →    + − −   + −  =  + =  +  + +       2 2 1 sin lim 2 2 1 x x x x →     =  +   = + =   + +   38) 2 2( ) 0 1 cos cos cos cos cos lim lim x x x x x x x x x → → − + − (85)( ) ( ) ( ) 2 2 2 2 0 2sin cos cos cos 1 cos 2 1 cos 2 lim lim 1 cos x x x x x x x x x x x x x → →    − −   −    = + =  +     +      ( ) 2 2 0 sin sin 1 2 2 sin lim lim 2 cos 2 cos 2 x x x x x x x x x → →                   =   =     = + = +     +                Bài Tính giới hạn sau: 1) 0 cos cos lim cos cos x x x x x → − − ĐS: 1 3 2) 6 1 2sin lim 4 cos x x x  → − − ĐS: 1 3) 2 1 sin cos lim cos x x x x  → + + ĐS: 24) 0 sin sin lim sin x x x x → − ĐS: 5) 4 2 cos lim sin x x x   → −  −      ĐS: 2 6) 3 0 1 cos lim sin x x x → − ĐS: 7) 3 3 sin cos lim sin x x x x  → − ĐS: − 8) 4 lim tan tan x x x   →   −         ĐS: 9) 3 cos cos 2 lim sin x x x x  → + + ĐS: 3 10) 3 tan lim 2sin x x x  → − − ĐS: 1 12 − Lời giải 1) 0 0 sin cos cos 2sin sin sin lim lim lim lim sin cos cos 2sin sin sin 3 3 x x x x x x x x x x x x x x x x x x → → → → − = − = = = − −  2) 2 2 6 6 1 2sin 2sin 1 lim lim lim 4 cos 4sin 2sin x x x x x x x x    → → → − = − = = − − + 3) ( ) 2 2 2 1 sin cos 2 cos sin lim lim lim cos 2sin cos cos x x x x x x x x x x x    → → → + + = + = + = 4) 0 0 sin sin 2cos sin lim lim lim 2cos sin sin x x x x x x x x x x → → → − = = = (86)Page 86 5) 4 4 2 2 cos 2 cos cos 2 2 cos lim lim lim sin sin sin 4 4 x x x x x x x x x        → → →     −    −  − =   =    −   −   −              4 4 sin sin 8 lim 2 sin cos 2 8 x x x x x      →     −  +   −      =  −   −          2sin 8 lim cos x x x    →  +      = =  −      6) ( ) 2 2 0 2 3 3 2 1 cos cos cos lim lim tan sin cos cos x x x x x x x x x → → −  − =    + +    ( ) 2 0 3 3 2 cos lim 6 cos cos cos x x x x x → = =   +  + +    7) ( ) sin 3 3 3 2sin 3 sin 2sin 3 sin cos lim lim lim lim sin sin 3 sin 3 3 x x x x x x x x x x x x x x                → → → − −  −  →       −             −  −       − =   = =    = − − +   −         −   −      8) ( )2 2 4 4 2 tan tan tan lim tan tan lim lim 4 tan tan tan x x x x x x x x x x x     → → →    − =   − = =       − +  +   9) 3 3 3 cos cos 2 cos 3cos cos lim lim sin 3sin 4sin x x x x x x x x x x   → → + + = − + − ( )( ) ( )( ) 3 2 cos cos cos lim 2sin 2sin sin x x x x x x x  → − + = + − ( ) 3 cos cos cos cos lim 3 x x x x   →  −  +     = ( ) ( ) 3 sin cos cos 2 lim 3 cos 2sin sin 2 x x x x x x x    →   −  +  +   = =  +  +     10) ( ) ( ) ( ) 2 2 2 3 4 tan cos tan lim lim 2sin 1 tan . tan tan 1 x x x x x x x x x   → → − − = − −  + +      ( ) ( ) 2 3 4 cos lim 1 tan tan tan x x x x x  → − =   +  + +    1 12 (87)Bài 10 Tính giới hạn sau: 1) 0 cos lim sin x x x → − ĐS: 2) 0 1 sin cos lim 1 sin cos x x x x x → + − − − ĐS: 1− 3) 0 sin lim 1 sin cos x x x x → − − ĐS: 1− 4) 1 cos lim sin x x x → − ĐS: 5) 0 sin sin lim sin x x x x → − ĐS: 6) 0 1 lim sin tan xx x  −      ĐS: 7) 2 4 2 sin lim 2 cos x x x  → − − ĐS: 2 − 8) 6 sin lim 1 2sin x x x   →  −      − ĐS: 2 3 9) 4 sin lim 1 sin x x x   →  −      − ĐS: 10) 2 lim cot sin xx x  −   
- Xem thêm -

Xem thêm: Đề thi thử THPT quốc gia, Đề thi thử THPT quốc gia