Title 0305-bai-tap-ap-dung-cac-quy-tac-tim-gioi-han-ham-so-5595e842-fda9-48b6-b4a9-da93d8e7789c.pdf ✅

Học online tại: https://mapstudy.vn _________________________________________________________________________________________________________________________________________________________________________________________________________________ _________________________________________________________________________________________________________________________________________________________________________________________________________________ Thầy Phạm Ngọc Lam Trường 1 BÀI TẬP: GIẢI TÍCH I CHƯƠNG III: GIỚI HẠN VÀ SỰ LIÊN TỤC CỦA HÀM SỐ GIẢI BÀI TOÁN TÌM GIỚI HẠN Bài 1: CMR khi x 0 → 1) ( ) 2 2 a x 1 cosax a 0 2 −  2) ( ) 2 sinax sin bx ax a 0 +  3) ( ) x a 1 xlna 1 a 0 −   4) ln 1 ax ax a 0 ( +  ) ( ) 5) ( ) ( ) α 1 kx 1 k + −  αx kα 0 6) ( ) α α 1 α n α 1 n ax a x ... a x ax α 0,a 0 + + + + +   Hướng dẫn giải 1) Ta có: 2 2 2 2 ax ax a x 1 cosax 2sin 2. 2 2 2   − = =     khi x 0 → , đpcm. 2) ( ) 2 2 2 2 x 0 x 0 x 0 x 0 x 0 sinax sin bx sinax sin bx sinax sin bx bx lim lim lim lim 1 lim 1 0 1 → → → → → ax ax ax ax ax ax +   = + = + = + = + =     2  + sinax sin bx ax , đpcm. 3) x x L x x 0 x 0 x 0 a 1 0 a lna lim lim lima 1 → → → xlna 0 lna −    = = =    đpcm. 4) ( ) ( ) L x 0 x 0 x 0 ln 1 ax a / 1 ax 0 1 lim lim lim 1 → → → ax 0 a 1 ax + +    = = =    + đpcm. 5) ( ) ( ) ( ) α α 1 L α 1 x 0 x 0 x 0 1 kx 1 0 α 1 kx k lim lim lim 1 kx 1 kαx 0 ka − − → → → + − +    = = + =    đpcm. 6) ( ) ( ) ( ( )) α α 1 α n α 1 α α n 1 L 1 n 1 n α α 1 x 0 x 0 x 0 ax a x ... a x 0 aαx a α 1 x ... a α n x lim lim lim 1 o x 1 ax a 0 αx + + − + − → → → − + + +   + + + + + = = + =     α α 1 α n α 1 n ax a x ... a x ax  + + + + + , đpcm. Bài 2: Khi x 0 → , cặp VCB sau có tương đương không? NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi Đăng Ký Khóa Học Online Tại Fanpage: Tài Liệu Khóa Học Mappi Tài Liệu Được Chia Sẻ Bởi Fanpage: Tài Liệu Khóa Học Mappi
Học online tại: https://mapstudy.vn _________________________________________________________________________________________________________________________________________________________________________________________________________________ _________________________________________________________________________________________________________________________________________________________________________________________________________________ Thầy Phạm Ngọc Lam Trường 2 1) α x x x ( ) = + , ( ) sinx β x e cos x = − 2) α x ln cos x ( ) = ( ), ( ) 2 tan x β x 2 = − 3) α x arctan sin2x ( ) = ( ) , ( ) tanx β x e cos 2x = − 4) ( ) 3 α x x x = − , β x cos x 1 ( ) = − Hướng dẫn giải 1) sinx sinx sinx sinx 1/4 1/4 1/4 1/4 x 0 x 0 x 0 x 0 x 0 e cos x e 1 1 cos x e 1 1 cos x e 1 1 cos x lim lim lim lim lim x x x x x x x → → → → → − − + − − − − −   = = + = +   +   ( ) ( ( )) 2 7/4 7/4 3/4 1/4 1/4 1/4 x 0 x 0 x 0 x 0 x 0 x 0 sinx x / 2 x x x lim lim lim l 0 im limx lim β x x x 2 2 α x x o → → → → → → = + + =  = + = = 2) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 x 0 x 0 x 0 x 0 x 0 x x ln 1 2sin 2sin ln cos x ln cos x x / 2 2 2 lim 2lim 2lim 2lim 4lim 1 tan x tan x tan x x x 2 → → → → →     − −   = − = − = − = = − α x ~ β x ( ) ( ) 3) ( ) ( ) tanx tanx 2 2 x 0 x 0 x 0 x 0 x 0 x 0 e cos 2x e 1 1 cos 2x tanx 2sin x x 2x 1 lim lim lim lim lim lim → → → → → → arctan sin2x arctan sin2x sin2x sin2x 2x 2x 2 − − + − = = + = + = α x , β x ( ) ( ) là các VCB cùng bậc 4) ( ) ( ( )) 2 5/3 3 3 1/3 x 0 x 0 x 0 x 0 x cosx 1 1 cosx x / 2 1 lim lim lim limx 0 x x x x x 2 β x o α → → → → − − = = = − = −  = − − Bài 3: So sánh cặp VCB sau đây: 1) Khi x 0 → + : ( ) 3 4 α x x x x = − − , β x 1 x 1 ( ) = − − 2) Khi x→+ : ( ) 2 1 1 α x x x = + , ( ) 2 2 x 1 β x ln x + = 3) Khi x 0 → : ( ) 3 2 3 α x x x = + , ( ) sinx β x e 1 = − 4) Khi x 0 → : α x tanx ( ) = , ( ) sinx 2 β x e x 1 = − − 5) Khi x 0 → : ( ) 3 2 α x 3x 2x = − , β x ln cos 2x ( ) = ( ) 6) Khi x 0 → + : ( ) 3 2 α x x x = + , ( ) sinx β x e cos 2x = − NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi Đăng Ký Khóa Học Online Tại Fanpage: Tài Liệu Khóa Học Mappi Tài Liệu Được Chia Sẻ Bởi Fanpage: Tài Liệu Khóa Học Mappi
Học online tại: https://mapstudy.vn _________________________________________________________________________________________________________________________________________________________________________________________________________________ _________________________________________________________________________________________________________________________________________________________________________________________________________________ Thầy Phạm Ngọc Lam Trường 3 Hướng dẫn giải 1) Đáp án: β x o α x ( ) = ( ( )) (Gợi ý: 3 4 1/4 x x x ~ x − − − ) 2) 2 2 2 x x 2 2 x 1 1 ln ln 1 x x L lim lim 1 1 1 1 x x x x →+ →+ +   +     = = + + Đặt: ( ) ( ) ( ( )) 2 2 2 t 0 t 0 t 0 ln 1 t 1 x t t L lim lim lim t 0 x t t β x o α t → → → + + + + = = = = =   = + 3) ( ) = + 2 3 2 3 3 α x x x x ( ) = − sinx β x e 1 sinx x  = β x o α x ( ) ( ( )) 4) sinx 2 sinx 2 2 x 0 x 0 x 0 x 0 x 0 e x 1 e 1 x sinx x lim lim lim lim lim 1 → → → → → tanx tanx tanx x x − − − = − = − =  α x ~ β x ( ) ( ) 5) ( ) = − − 2 2 2 α x 3x 2x 2x ( ) ( ) ( ) 2 2 2 β x ln cos 2x ln 1 2sin x 2sin x 2x = = − − −  α x ~ β x ( ) ( ) 6) ( ) = +3 2 α x x x x ( ) = − = − + sinx sinx 2 β x e cos 2x e 1 2sin x ( ) ( ) sinx 2 2 x 0 x 0 x 0 x 0 x 0 β x e 1 2sin x sinx 2x lim lim lim lim lim 1 → → → → → α x x x x x − = + = + =  α x ~ β x ( ) ( ) Bài 4: Với α x ,β x ,γ x ( ) ( ) ( ) là các VCB trong cùng một quá trình x a, → hãy chứng minh: a) Nếu α x β x ( ) ( ) và β x γ x ( ) ( ) thì α x γ x ( ) ( ). b) Nếu α x β x ( ) ( ) và β x o γ x ( ) =   ( )   thì α x o γ x ( ) =   ( )   . Hướng dẫn giải a) ( ) ( ) ( ) ( ) x a x a α x α x lim 1 lim 1 → → β x γ x =  = (Vì β x γ x ( ) ( ) ) α x γ x ( ) ( ) , đpcm. NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi Đăng Ký Khóa Học Online Tại Fanpage: Tài Liệu Khóa Học Mappi Tài Liệu Được Chia Sẻ Bởi Fanpage: Tài Liệu Khóa Học Mappi
Học online tại: https://mapstudy.vn _________________________________________________________________________________________________________________________________________________________________________________________________________________ _________________________________________________________________________________________________________________________________________________________________________________________________________________ Thầy Phạm Ngọc Lam Trường 4 b) ( ) ( ) ( ) ( ) ( ) ( ) x a x a β x α x β x o γ x lim 0 lim 0 → → γ x γ x =  =  =     (Vì α x β x ( ) ( ) )  = α x o γ x ( )   ( )   , đpcm. Bài 5: Tính các giới hạn sau: 1) 2 x 3 x 3x 1 5x lim →− 2x 1 + + − + 2) 3 x 0 2x 1 1 lim x 1 1 → + − + − 3) ( ) 2 x lim x 2x x →− − + 4) ( ) 2 2 x lim x 4x 1 x x 5 → + − − + − 5) 1 2 x 2 x 5x x 1 lim → 4x x 1   + +     − + 6) ( ) ( ) 3 2 x 0 1 5x sin x 1 lim → tan x ln cos 3x − − − Hướng dẫn giải 1) ( ) 2 2 x 3 16 5 x 3x 1 5x 3 3. 3 1 5. 3 lim →− 2x 1 2. 3 1 + + − + − + − − + − + − − = = 2) ( )( ) ( ) ( ) ( ) ( ) 3 x 0 x 0 x 0 2 2 3 3 3 3 2x 1 1 3 x 1 1 2 x 1 1 2x 1 1 lim lim lim x 1 1 x 1 1 2x 1 2x 1 1 2x 1 2x 1 1 4 → → → + − + + + + + − = = = + −     + − + + + + + + + +         3) ( ) →− →− − − + = = − − 2 x x 2 2x lim x 2x x lim 1 x 2x x 4) ( ) 2 2 x 2 lim x 4x 1 x 3 x 5 → + − − + − = (Gợi ý: nhân liên hợp) 5) 2 2 2 2 x x 5x x 1 5 1/x 1/x 1 ln ln 2 4x x 1 4 1/x 1/x x lim lim x x 0 2 x 1 5x x 1 lim e e e 4x x 1 → →     + + + +             − + − + →   + +   = = = =   − + (Chú ý: ( ) ( ) ( ) ( ) x x0 0 lim B x lnA x B x x x lim A x e → → = ) 6) ( ) ( ) ( ) ( ) ( ) 2 3 2 x 0 x 0 2 2 2 2 3 3 1 5x sin x 1 1 5x sin x 1 lim lim tan x ln cos 3x 3x x .ln 1 2sin 1 5x sin x 1 5x sin x 1 2 → → − − − − = −    − − − + − +       ( ) 2 x 0 2 2 2 2 3 3 5 sin x lim 3x ln 1 2sin 1 5xsin x 1 5xsin x 1 2 → =    − − + − +       ( ) 2 x 0 2 2 2 2 3 3 5 x lim 3x 2sin 1 5xsin x 1 5xsin x 1 2 → =   − − + − +     NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi NNNNNN https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi https://www.facebook.com/tailieukhoahocmappi Đăng Ký Khóa Học Online Tại Fanpage: Tài Liệu Khóa Học Mappi Tài Liệu Được Chia Sẻ Bởi Fanpage: Tài Liệu Khóa Học Mappi