TÀI LI U B I D NG CHUYÊN MÔN L P 11 CH NG IV GI I H N ng V n ông –Tr GV: D ng THPT Tân Yên Gi i thi u đ n em b i Tr n Qu c Hoài http://bsquochoai.ga I Gi i h n c a dãy s Gi i h n h u h n Gi i h n đ c bi t: 1 lim  ; lim  (k   ) k n n n n Gi i h n vô c c Gi i h n đ c bi t: lim n  n ) lim qn   (q  1) lim q n  ( q  1) ; lim C  C n lim nk   (k  n   n 2 nh ệí : a) N u lim un = a, lim = b  lim (un + vn) = a + b  lim (un – vn) = a – b  lim (un.vn) = a.b u a  lim n  (n u b  0) b n nh ệí: a)N u lim un   lim 0 un b) N u lim un = a, lim =  lim c) N u lim un = a  0, lim = un  neáu a.vn  lim =  b) N u un  0, n lim un= a a  lim  neáu a.vn  un  a d) N u lim un = + , lim = a un =0   neáu a   neáu a  c) N u un  ,n lim = lim(un.vn) = lim un = d) N u lim un = a lim un  a * Khi tính gi i h n có m t d ng vô đ nh: T ng c a c p s nhân ệùi ốô h n  u1 ,  – , 0. ph i tìm cách kh d ng vơ đ nh S = u1 + u1q + u1q + … =  q  1  1 q M t s ph ng pháp tìm gi i h n c a dãy s :  Chia c t ốà m u cho ệu th a cao nh t c a n 1 n 1 n 1 VD: a) lim  lim 2n  2 n 1  n  n  3n n  lim 1 b) lim 1  2n 2 n   c) lim(n2  4n  1)  lim n2        n n2   Nhân ệ ng ệiên h p: Dùng h ng đ ng th c  a  b  a  b   a  b;   VD: lim n2  3n  n = lim   a  b   a2  ab  b2   a  b n2  3n  n   n2  3n  n n2  3n  n   = lim 3n n2  3n  n ThuVienDeThi.com = ,  Dùng đ nh ệí Ệ p: N u un  ,n lim = VD: a) Tính lim lim un = sin n n sin n 1 sin n  lim  nên lim 0 n n n n 3sin n  cos n b) Tính lim 2n  Vì  Vì 3sin n  cos n  (32  42 )(sin2 n  cos2 n)  nên  3sin n  cos n  2n2  3sin n  cos n Mà lim  nên lim 0 2n2  2n2  2n2  Khi tính gi i h n d ng phân th c, ta ý m t s tr ng h p sau đây:  N u b c c a t nh h n b c c a m u k t qu c a gi i h n b ng  N u b c c a t b ng b c c a m u k t qu c a gi i h n b ng t s h s c a lu th a cao nh t c a t c a m u  N u b c c a t l n h n b c c a m u k t qu c a gi i h n + n u h s cao nh t c a t m u d u k t qu – n u h s cao nh t c a t m u trái d u(ta th ng đ t nhân t chung c a t , m u riêng) Baøi 1: Tính gi i h n sau: (Chia c t m u cho na v i s m a cao nh t Ho c đ t nhân t chung) 1) lim(n2  n + 1) S: + 3n3  2n2  n lim S: 13) 2) lim(n + n + 1) S: - n  3) lim 2n  3n  S: + n4 3 14) lim S: 4) lim  2n  n S: - ( n  1)(2  n )( n  1) 5) lim(2n + cosn) S: + – n2 + n – 1 15) lim S: -1/2 6) lim( n  3sin2n + 5) S: + 2n2 – 4n – 3n  16)lim S: S: + 7) un = n n+1 1 2n  8) un = 2n  3n S: -  17)lim S: n  2n  2n  9) lim S: 2n  n2  n3  4n2  S: + 18) lim n2  3n3  2n2  10) lim S: 3n3  2n2  n 2n  n  lim S: - 19) n2  n  S: 11) lim 2n  n  4n2  2n  20) S: - lim 2n2  n  3 n  12) lim S: 2/3 3n2  2n  Bài 2: Tính gi i h n sau: (Chia cho l y th a có c s l n nh t) 1) lim 2) lim  3n  3n 4.3n  7n1 2.5n  7n S: 3) lim S: 4) lim 4n1  6n2 5n  8n 2n  5n1 ThuVienDeThi.com  5n S: S: 5) lim  2.3n  7n S: -1/2 6) lim  2.3n  6n S: 1/3 5n  2.7n 2n (3n1  5) Bài 3: Tính gi i h n sau: (T d ng vô ±vô cùng; M u d ng vô + vô ;b c c a t m u b ng ta chia cho s m cao nh t;) k k k Chú ý: n k có m ; n có m 4n   n  1) lim n2   n  2) lim 3) lim S: n  4n   n S: n 2 n n2   n6 S: 4n2   n 4) lim 5) lim 6) S: n2  4n   n (2n n  1)( n  3) (n  1)(n  2) lim S: n  4n  4n  S: -1/(  ) 3n2   n n   n2 Baøi 4: Tính gi i h n sau: N u tốn có d ng: + Vơ – vơ khơng có m u (h s c a n b c cao nh t gi ng nhau) + C t m u d ng: Vô cùng- vô (h s c a b c cao nh t gi ng nhau) Thì ta nhân liên h p có c n b c 2,3 r i chia cho l y th a có s m cao nh t N u tốn d ng vơ + vơ kq vô ta đ t nhân t chung có m cao nh t r i tính gi i h n Ho c h s c a n b c cao nh t khác ta chia ho c đ t nhân t chung 1) lim( n2  3n  n) S: + 9) lim 1  n2  n4  3n   S:   2) lim( n2  2n  n  2013) S: 2012 n  4n  4n  10) lim S: -1/(  ) S: -1/2 3) lim n  n  n 3n2   n 4) lim( n2   n  5) S: 11) lim S: - 5) lim( n2  2013  n  5) S: n2   n2  4n   n  6) lim  n2  2n  n  1 S: 12) lim S: -1/2   n  n   n 7) lim  n2  n  n2   S: 1/2   n2   n6 S: 13) lim 8) lim  2n  n3  n  1 S: -1 n   n2   Baøi 5: Tính gi i h n sau: (Gi i h n k p gi a hai bi u th c có k t qu )  1) lim  cos n2 n2  S: 3) lim 3sin6 n  5cos2 (n  1) n2  3sin (n3  2)  n2 (1)n sin(3n  n2 ) lim 4) 2) lim S:  3n2 3n  Bài 6: Tính gi i h n sau: (Rút g n r i tính gi i h n)   1    n 5) lim    1) lim   S: 1/2 (2n  1)(2n  1)  n2  3n  1.3 3.5  1     2) lim  S: 3/2  n(n  2)   1.3 2.4    1  3) lim         S: 1/2  22  32   n2   1    4) lim    S: n(n  1)   1.2 2.3 6) lim   22   n   32   3n ThuVienDeThi.com S: S: S: -1/3 S: 1/2   1   Baøi 7: Cho dãy s (un) v i un =      1   ,v i n 2     n2  a) Rút g n un S: (n+1)/2n b) Tìm lim un S: 1/2 1   Baøi 8: a) Ch ng minh: (n  N*) n n   (n  1) n n n 1 1    b) Rút g n: un = 2 3 n n   (n  1) n c) Tìm lim un S : u1   Baøi 9: Cho dãy s (un) đ c xác đ nh b i:     ( u 1) u n n  n   2n a) t = un+ – un Tính v1 + v2 + … + theo n b) Tính un theo n c) Tìm lim un S: u  0; u2  Baøi 10: Cho dãy s (un) đ c xác đ nh b i:  2un2  un1  un , (n  1) a) Ch ng minh r ng: un+ =  un  , n  2 b) t = un – Tính theo n T tìm lim un S: 2/3 u1  2012  u u u Cho dãy s (un) xác đ nh b i  ; nN* Tìm lim (    n ) (HSG l ng s n 2011) n  u u3 u n 1 u n 1  2012.u n  u n S: - CM đ c dãy t ng : u n 1  u n  2012u n2  n - gi s có gi i h n a : a  2012a  a  a   2012 Vô Lý nên limun =  un u n2 (u  u n ) 1 - ta có :  (   n 1 )  u n 1 u n 1u n 2012u n 1u n 2012 u n u n 1 1 1 V y: S lim(  ) 2012 n  u1 u n 1 20122 Baøi 11: Cho dãy (xn) xác đ nh nh sau:  x1  ( n  N *)  x x 3x    n n  n 1 1 ( n  N * ) Tìm LimSn (HSG l ng s n 2012) t Sn     x1  x  xn  Baøi 12: T ng Dãy c p s nhân lùi vô h n: (1)n 1    n 1  b S = + S: a b.12/11 10 102 10 Baøi 13: Bi u di n s th p phân vô h n tu n hoàn sau d i d ng phân s : 1 a S = + + + … a 0,444 b 0,2121  a  a   a , v i a, b < b  b   b n Baøi 14: L = lim n n   II Gi i h n c a hàm s Gi i h n h u h n Gi i h n đ c bi t: c 0,32111 S: a.4/9 b.21/99 c.289/900 S: (1-b)/(1-a) Gi i h n vô c c, gi i h n Gi i h n đ c bi t: ThuVienDeThi.com vô c c lim x  x0 ;  k chẵn lim x k   ; lim x k   x  x   k lẻ x  x0 lim c  c (c: h ng s ) x  x0 nh ệí:  lim f ( x )  L  x  x0 a) N u  lim g( x )  M   x  x0 thì: * lim  f ( x )  g( x )  L  M lim c  c ; x  x0 * lim  f ( x )  g( x )  L  M x  x0 * lim  f ( x ).g( x )  L.M x  x0 f ( x) L (n u M  0)  x  x0 g( x ) M f(x)  b) N u  lim f ( x )  L  x  x0 * lim * L  * lim x  x0 f ( x)  L c) N u lim f ( x )  L x  x0 lim f ( x )  L x  x0 Gi i h n m t bên: lim f ( x )  L c 0 xk lim   x 0 x x    ; x 0 x 1 lim  lim   x 0 x x 0 x nh ệí:  lim f ( x )  L   x  x0 a) N u  thì: lim g( x )     x  x0  neáu L lim g( x )   x  x0 * lim f ( x )g( x )   lim g( x )  neá u L  x  x0  x  x0 f ( x) * lim 0 x  x0 g( x )  lim f ( x )  L   x  x0 b) N u  thì: lim g( x )    x  x0 lim lim x  x0  f ( x )  neáu L g( x )   g( x )  L.g( x )  Khi tính gi i h n có m t d ng vô đ nh: x  x0  lim  f ( x)  lim  f ( x)  L x x0 x  lim 0. ph i tìm cách kh d ng vơ đ nh x  x0 M t s ph ng pháp Ệh d ng ốô đ nh: D ng P( x ) a) L = lim ố i P(ồ), Q(ồ) ệà đa th c ốà P(ồ0) = Q(x0)= x  x0 Q( x ) Phân tích c t m u thành nhân t rút g n x3  ( x  2)( x  x  4) x  x  12  lim  lim  3 VD: lim x 2 x  x 2 x 2 ( x  2)( x  2) x2 P( x ) b) L = lim ố i P(ồ0) = Q(x0) = ốà P(ồ), Q(ồ) ệà bi u th c ch a c n b c x  x0 Q( x ) S d ng h ng đ ng th c đ nhân l ng liên h p t m u    x    x  2 4 x 1  lim  lim  x 0 x 0 x 0   x x x 2   x  P( x ) c) L = lim ố i P(ồ0) = Q(x0) = ốà P(ồ) ệà biêu th c ch a c n Ệhông đ ng b c x  x0 Q( x ) VD: lim Gi s : P(x) = m u( x )  n Ta phân tích P(x) = VD: lim x 0 v( x ) với m u( x 0)  n v( x0 )  a  m u( x)  a    a  n v( x)   x 1 1 1 1 x  x 1  1 x  lim    x 0  x x x  ThuVienDeThi.com  , ,  – ,    1 1 = lim      x 0  3  1   x  ( x  1)  x     P( x ) D ng : L = lim ố i P(ồ), Q(ồ) ệà đa th c ho c bi u th c ch a c n x  Q( x )  – N u P(x), Q(x) đa th c chia c t m u cho lu th a cao nh t c a x – N u P(x), Q(x) có ch a c n có th chia c t m u cho lu th a cao nh t c a x ho c nhân l h p 2  2 x  5x  x x2 VD: a) lim  lim 2 x  x  x  x  1  x x2 2x  b) lim x  x 1  x  lim x  2  1 1 x2 D ng  – : Gi i h n th ng có ch a c n Ta th ng s d ng ph ng pháp nhân l ng liên h p c a t m u VD: lim x    x  x   lim   x  x   x  x  x  D ng 0.: Ta c ng th ng s d ng ph VD: lim ( x  2) x 2 x x2   1 x ng liên  lim x 2 1 x  x  lim x  ng pháp nh d ng x  x x2  1 x  x 0 0 Bài 1: Tìm gi i h n sau: + Khi thay x=a vào f(x) th y m u khác gi i h n b ng f(a) + Khi thay x=a vào f(x) th y m u b ng t khác gi i h n b ng  1) lim (x2 + x) S: 12 x2  x  x 3 7) lim S: x x 2  x S: ± 2) lim x 1 x  x2  2x  3 S: / 8) lim 1 x  x  x x 1 x 1 S: 3) lim x 0 1 x x 8 3 9) lim S: 3x   x  x  x 4) lim S: -3/2 x 1 x 1 3x   3x  10) lim S:   sin  x   x 2 x 1  S: /   5) lim  x 11) lim x sin S: x x 0 2 lim x 1 S:-2/3 x4  x  Bài 2: Tìm gi i h n sau: (Khi thay x=a vào f(x) th y t =0; m u =0 ta rút g n m t nhân t r i thay ti p t i m u khác xong) cịn n u m u =0 t khác kq  6) x 1 x2 1 S: x 1 x  1  2) lim x    S: -1 x0  x 1) lim 3) lim x2 x3  x2  4) lim x 1 3x  x  x 1 S: 2x  3x  S: x 2 x2 5) lim S: ThuVienDeThi.com 6) lim x  16 S: -8 x3  x2 x3  x2  x  x 2 7) lim S: x  3x  x  3x  5x  8) lim S:1 x 1 x 1  x  x2  x3 9) lim S: x1 1 x x  5x  3x  S: 10) lim x 3 x  8x  x5  11) lim S: 5/3 x 1 x  x 1 12) lim x  5x  x S: 10 (1  x )2 x  5x  x S: 13) lim x 1 x 1    14) lim   S: -1/2 x 1 x  x 1   Bài 3: Tìm gi i h n sau: (M t c n b c 2) x 1 1) lim x 2 4x   S:1/6 x 4  x2 1 S:0 2) lim x 0 x x 5 3 4x 3) lim x 4 x 3 4) lim x 9 9x  x2 S: -1/6 S:-1/54 Bài 4: Tìm gi i h n sau: (Hai c n B c 2) 1 x  1 x 1) lim S: x 0 x x 1 S:2 2) lim x 1 x3 2 3) lim x   x x 2 4x   4) lim x 2 5) lim x 2 2 x 7 3 2x   2 x3 x2  x 6) lim S:3 x 1 x 1 x 1 7) lim   x x 4 1  x   15) lim  S: -1  x 1   x  x    x2  x4 16) lim    S: x 1 x  5x  3(x  3x  2)   x1992  x  S: 1993/1992 x 1 x1990  x  xm 1 18) lim ý t ng c a CSN S: m/n x 1 x n  (1  5x )(1  x )  S: 14 lim x 0 x (1  x )(1  x )(1  3x )  19) lim S: x 0 x 17) lim x  x   x n  n S: n(n+1)/2 x 1 x 1 x n  nx  n  21) lim S: n(n-1)/2 x 1 (x  1)2 20) lim 5) lim x 7 2 x 3 x  49 2x   x  S: -4/15 x 1 x  4x  x  3x  lim 7) S: 9/4 x 1 x2 1 6) lim 8) lim x   x  3x S:1/2 x 1 9) lim 2x   x  3x  x 1 x  1  x 1 11) lim S:3/2 12) S:-4/3 13) lim x   3x  8) lim S:-1/4 x 1 x 1 x 0 x 1 1  2x  lim x 2 x 0 14) lim x 3 15) lim x 0 ThuVienDeThi.com S:1/6 x   x  S: x 1 10) lim S:-3/4 S:-1/3 S: -1/56 S:-3/4 x   2x x 1   x x2   x  16  x   2x x  3x S:-1/4 S:4 S:-2/9 x   x  16  S: 7/24 x x  a  xa 16) lim x a x a 2 , v i a> S: 1/ 2a x 1 17) lim x 1 S:2 x   x  3x Bài 5: Tìm gi i h n sau: (M t c n B c 3) 4x  S :1/3 x 2 x2 2x   2) lim S:2/3 x 1 x 1 1 x 1 5) lim x 0 x2 x 1 6) lim x1 4x   3 1) lim x 3) lim 1 x 1 x5  x3  x 0 S:3 5x   x 7) lim x 0 S:1/3 S:1 S:1 S:24 x 1 Baøi 6: Tìm gi i h n sau: (Hai c n khác b c) 4) lim x 1 1 x  1 x x 1) lim x 0 S:4/3 13) lim 2x   x  x 0 3) 12) lim x 1  x 1 2) lim S :1/6 1 x 1 lim x 0  x x 2 x4 x S:-1/18 x  x  5x  x  10  x  S:-7/72 6) lim x  3 x2  1 4x  1 6x 7) lim S:0 x x 10  x  x  8) lim S:-1/3 x 2 x2 (1  x )(1  x )(1  x )(1  x ) S:1/120 x 1 (1  x)4 17) lim 18) lim x 0 19) lim S:7/54 Bài 7: Tìm gi i h n sau: ( lim x 0 sin x  x x 1) lim S: 2/ S:1 2) lim x  cos x tan x  s in2x S: 3) lim x 0 cos x tgx 4) lim S:4/3  x x x 1  1 x x x x 0  x x  3x   x2   x2 10) lim S:2 x0 x2 x  11  x  11) lim S:7/162 x 2 x  5x  x 2 S:-1/24 14) lim x  11  x  x6  x2 x2  S:-11/24  4x  6x 1 S:5 x 0 x  x  x  15) lim S:7/3 x x (1  n x ) S: 1/n 16) lim x 1 (1  x) S:3/2 5) lim x2  x 1 1 1 x   x S:13/12 4) lim x0 x 9) lim  x3  x2  8) lim x 1  8 x x   x  3x  x   x2  x  sin x ta n x =1)  ; lim x  x x sin 5x x  3x 5) lim S:5/3 6) lim sin 5x sin 33 x sin x S:1/3 x0 45 x  cos 2x S:2 x0 x sin x  cos 4x 8) lim S:4 x 0 2x2 7) lim ThuVienDeThi.com S:5/6 S:-6 S:0 sin 2x 9) lim x 1 1  cos2x x 0 10) lim x0 11) lim x0 12) lim S:4 S: x2 cos x  cos 7x S:12 x2 cos x  cos3x S:2 sin x x0 sin x x 0 tan 2x 13) lim 15) lim x x0 17) lim x0 18) lim x x sin  cos x S:3 S:0  cos3x S:9/25 x0  cos5x  cos 2x xsin x 21) lim x 0 sin 2x  tan3x x 0 x 23) lim x 0 S:1 S:5 S: -1 S:16/ sin( x  1) S:-1/2 x2  4x    sin   x   S:1  39) lim  sin x x  x1 sin x  S:-1/2 cos x   sin x  cos x  tgx  x 42) lim  tgx  x   cot gx x 2 S: S:  x 8 tan( x  2) S:12    45) lim   x S: x   sin x sin 3x   sin 2x  cos 2x S:-1 x 0  sin 2x  cos 2x tan(a  x).tan(a  x)  tan a S:tan4a-1 46) L  lim x 0 x2 22) lim 47) lim (a  x)sin(a  x)  a sin a x x 49) lim x   S:0 x 50) lim x® 51) lim x2  sin x - + cos x tan x  sin2 x  cos x x0 ThuVienDeThi.com sin2 x S: (a+1)sina  2x   sin x x 0 30) lim tan x tan    x  S: 1/2  x  4 2 S: -1 43) lim ( x sin  ) x  x 44) lim  48) ( HGTVT-98): lim px cos cos x  29) lim x 1 1 x lim x® S:0 cos x  tan x 37) lim  41) lim  sin x  cos x sin x 28) lim - x sin x  S:-1/2 cos x  x tan x  sin x S:1/2 x 0 x3 cos4x  cos3x.cos5x 25) lim S: 18 x 0 x2  cos( cos x) S: B góc ph chéo 26) lim x 0 x sin sin 3x S: t n ph 27) lim x   cos x 24)  x S:1/  sin x 40) lim S:4 sin x  sin x 3sin x 22) lim x 38) lim 19) lim S:0 sin   x 35) lim x x  cos x  cos x x 0  36) lim S:1/9   sin 3x 20) lim 34) lim (1  cos 2x ) tgx  16) lim sin x cos x  sin x S:0 x0 tan( x  1) S:1/2  cos x sin 33) lim x   x S:-7/4 x1 x 14) lim  cos x cos x cos 3x S:14 x  tgx   x ) sin( x  S: -2 S:3 32) lim  x   sin x  x 31) lim 3x    x S:1 S: S:1 /8 S:0 52) lim (1- x)tan x® px S:2/ 3x2 - + x2 + S:4 x® 1- cos x x2 S:4/3 54) lim x® + x sin x cos x 53) lim + sin x - 1- sin x S:2 x® x cos x - cos x S:-1/12 56) lim x® sin x 57) lim 2sin2 x  sin x 1 S:-1 x 0 2sin x  3sin x  55) lim  cos x.cos 2x.cos3x S:7 x2  cos x.cos 2x.cos3x cos nx 59) lim x 0 x2 58) lim x 0   cos x  cos     60) lim x 0 sin  tan x  S:n(n+1)(2n+1)/12 S:0 61) lim  sin x   sin x x 0 tan x S:1  cot x S:-3/4 x   cot x  cot x 62) lim   cos x cos 2x cos3x x 0  cos 2x S:3/2 63) lim Bài 8: Tìm gi i h n sau: (gi ng gi i h n dãy s chia cho m cao nh t, nhân liên h p, giá tr t đ i) x 1) lim (3x3 5x2 + 7) S: - 18) xlim S:1 x   x  5  x 1 2) lim (2 x  3x) S:+  x 2x  7x  12 19) lim S: / 3 S:±  3) lim (2 x  3x) x  | x | 17 x  4) lim x  2x  3x  12 S:+  x  S:±  5) lim x  3x  x  x   22) lim x 10) 11) 12) S:+  lim  25) lim 3x(2x  1) lim x  (5x  1)(x x x 1 lim  2x) 14) lim x  15) lim x4  x  2x x  16) lim x  17) lim x   x  10 S:0 1  S: +  28) lim  S:-   x 2  x  x    S:1/3 4x    x S:-  S:-  x3  x2  x x 1 S:-2 x  x   3x  1  x4 1 27) lim S:4; -2/3 lim x2  S:1/2 2x2  x  2x2  x  S:-;+  30) lim x  x 2 2x2  31) lim S:0 x  x  x  29) x  3x  x 3x  (x  1)(x  3x  2) 26) lim    x 0 x x  S:+  x  x2  x x 1 S:6/5 x  x 1 4x  13) lim S:-2/3; 2/3 x  3x  x  2x   S:-  24) lim  x 1 (x  1)2 2x     S:-1/5 x    3x  5x S:-1;1 x2  x3  2x2  x 23) lim S:1 x  2x  x  2x S: ± 23) lim x  2 x  x  2x  S:2 x  x  x 1 S:-  8) lim 3x  2x 5x  x  S:- 2x  x  1  2x x2 21) lim x3  6) lim S:+  x  x  2x  x S:+  7) xlim  x  9) lim x  x4  x4 20) lim x  ThuVienDeThi.com t nhân t , d u x2  2x   4x  lim 32) x  4x    x x  3x  x lim 34) (2 x  1) x  x  5x x  38) lim x  4x  x  40) lim S:1 ( x  2) x  S:4 x   3x x   S:2/5 x  x  3x lim 35) S:3;1/5 x  x  11 S:+  x   2x  (1  x)(1  x)2 (3  x)2 39) lim S:1 x   (2  x)(3  x)2 (  x)2 x  S:0 S:-1;5 4x2  2x    x lim 33) x  x  10 37) lim 4x   x  2 x  5x  S:+  x  x  Bài 9: Tìm gi i h n sau: (gi ng gi i h n dãy s chia cho m cao nh t, nhân liên h p) 14) lim (3x   9x  12x  3) S:- ;0 1) lim  x  x  x  S:1/2 x    x    15) lim  x   x  x   S:0 2) lim ( x  x  x) S:+  x x    lim 36) 3) lim ( x2  3x   x) S:-3/2 x 5) lim x   x 1  x  x 17) lim ( x2  3x   x  2) S:-1/2 S:+  4) lim ( x2  3x   x) x S:0 18) lim ( x  3x   x  1) S:1/2;+  x  6) lim ( x  x   x) S:+ ;-1 19) lim 7) lim ( x   x  ) S:0 20) x x  x 2 8) lim ( x  4x   x  3x  2) S:1/2;-1/2 x 9) lim x   10) lim x   S:2 x2  x   x 2x   x 23) 11) lim x( x   x) S:-1/2; + x  12) lim x    x2   x 1  3x   x  25) lim x  26) lim 13) Cho f(x) = x  2x  - x  2x  Tính gi i h n lim f(x) lim f(x), t nh n x  lim x  S:0  x  x2  x  x  2x   2x  1 24) lim S:-1 lim x  S:+  x lim  x   x   S:0 x      21) lim  x  x  x  x  S:1/2 x    22)  S:+  16) lim ( x2  3x   x  2) x x xét v s t n t i c a gi i h n lim f(x) S :-2 ;2 x x 3  S:-   S:2 x2  x x   x  x  S:0   x  x S:2  x   x3  x   S:2/3 x  Bài 10: Tìm gi i h n sau: a lim x 1 x  b lim (  x  2x) c x5 S:a b 10 c.+ lim x 1 x x 1 d lim x 1 x x 1 1 x  x 1 e lim x 1 x2  x3 d - e Bài 11: Tìm gi i h n sau n u có S: a b -3 c.Ko xđ Bài 12: Tìm gi i h n sau: ( x  15 1) lim S:-  x 2 x  a lim x2 | 3x  | x2 b lim x2 | 3x  | x2 c lim ý đ n d u bi u th c t m u tính gi i h n này) x  15 2) lim S:+  x 2 x  ThuVienDeThi.com x2 | 3x  | x2 3) lim x 3  3x  x x 3 S:-  x2  S:+  x 2 x 2 2 x S:1/3 5) lim  x 2 x  x  2 x 6) lim S:-1/3 x 2 x  x  x2  2x S:0 7) lim x 2 x  3x  8) lim S:5/2 x 2 x 1 S:1 9) lim x 1 x  x 1 10) lim S:-1 x 1 x  4) 11) 12) lim lim x0  lim x2  x3 2x 2x x2  3x  S:-  x x2 x  3x  S:+  lim x x2 x 3 lim S:- ;+ x 4 x  x2  3x  S:+  lim  x  x  x  x2  3x  S:-  lim  x  x  x  13) lim 14) 15) 16) 17) lim 18) x 1  x  3x  x  5x  S: /3  1 x  lim  x  S:0;0 x  x0  19) 20) lim x 1  x2 x2 x 1 S:+ S:1/2 S:-1;1 4x  x Bài 13: Tìm gi i h n m t bên c a hàm s t i m đ c ch ra: (Gi i h n m t bên ti n t i s )   x2  taïi x  S:-6;-2; ko xđ 1) f ( x )   x  x  1  x x   x2  2x x     x3 taïi x  S:-1/6; 32; K xđ 2) f ( x )    x  16 x   x   x  3x  x    3) f ( x )   x  taïi x  S:-1/2; -1/2; -1/2 x  x     1 x 1 x   4) f ( x )    x  taïi x  S:3/2;3/2;3/2  x   Bài 14: Tìm giá tr c a m đ hàm s sau có gi i h n t i m đ c ch ra:  x3   taïi x  S:m=1 1) f ( x )   x  x  mx  x  x  m x   f ( x )   x  100 x  taïi x  S:m=1 x   2) x 3  x 0  ThuVienDeThi.com  x  3m x  1 f ( x)   taïi x  1 x  x  m  x  1  3) S: m=2  x    f ( x)   x  x3  taïi x  S:m=1;m=2 m2 x  3mx  x  4)  III Hàm s liên t c Hàm s ệiên t c t i m t m: y = f(x) liên t c t i x0  lim f ( x )  f ( x0 )  x  x0 xét tính liên t c c a hàm s y = f(x) t i m x0 ta th c hi n b c: B1: Tính f(x0) B2: Tính lim f ( x ) (trong nhi u tr ng h p ta c n tính lim f ( x ) , lim f ( x ) ) x  x0 x  x0  x  x0  B3: So sánh lim f ( x ) v i f(x0) rút k t lu n x  x0 Hàm s ệiên t c m t Ệho ng: y = f(x) liên t c t i m i m thu c kho ng Hàm s ệiên t c m t đo n [a; b]: y = f(x) liên t c kho ng lim f ( x)  f (a), lim f ( x)  f (b) x a (a; b) x b  Hàm s đa th c liên t c R  Hàm s phân th c, hàm s l ng giác liên t c t ng kho ng xác đ nh c a chúng Gi s y = f(x), y = g(x) liên t c t i m x0 Khi đó:  Các hàm s y = f(x) + g(x), y = f(x) – g(x), y = f(x).g(x) liên t c t i x0 f ( x)  Hàm s y = liên t c t i x0 n u g(x0)  g( x ) N u y = f(x) liên t c [a; b] f(a) f(b)< t n t i nh t m t s c  (a; b): f(c) = Nói cách khác: N u y = f(x) liên t c [a; b] f(a) f(b)< ph ng trình f(x) = có nh t m t nghi m c (a; b) M r ng: N u y = f(x) liên t c [a; b] t m = f ( x ) ,M = max f ( x ) Khi v i m i T  (m; M) t n t i  a;b  a;b nh t m t s c  (a; b) cho f(c) = T Bài 1: Xét tính liên t c c a hàm s t i m đ c ch ra: x 3   x  5x  x   x  taïi x  S:Lt 1) f ( x )   x  x  taïi x  1 S: LT 5) f ( x )   x  3x  1 x  1 x    x 3 2 x  3x  x  x   6) f(x) =  t i xo = S:K Lt  x taïi x  S:Lt 2) f ( x )   2x  x   1 x     x2  x3  x  x   x  7) f(x) =  x  t i xo = S:K Lt     x x 3) f(x) =  t i xo = S: Lt 1  2x khix   11   3 x  1  2x  x   t i xo = 4) f(x) =   x 1 x   S:Lt   x  x  t i xo = 8) f(x) =   x   x    x  ThuVienDeThi.com S: Lt 9)  x 5 x   f (x)   x   taïi x  S:Lt ( x  5)2  x    x 1  11) f ( x )    x  2 x  x  taïi x  S:Lt x  1  cos x x  10) f ( x )   taïi x  S:K Lt x   x 1 Bài 2: Tìm m, n,a đ hàm s liên t c t i m đ c ch ra:  x3  x2  x   1 x  1 x  x  x  taïi x  S:m=0  1) f ( x )   x x 1 5) f(x)= t i xo= S:a=-3  3 x  m x   x a  x   x   x  2x  x  t i x = S:a=5/2 2) f(x) =  x   3x   a x   x   x2 t i x0 = S:a=0 6) f(x)=   x2  x 1 ax + 3) f ( x )   taïi x  S:m=2 x   2mx  x  3x  2x  x  4) f(x) =  t i x0 = S:a=2 x  2x  a Bài 3: Xét tính liên t c c a hàm s sau t p xác đ nh c a chúng:  x2  x  3x  x  2 1) f(x) =  Lt / R x   5) f ( x )   x  S: Lt / R x  2 1  x  x  2   x  3x  x   x  3x  10 2) f ( x )  5 x  S:K Lt t i x=2 x    x  2 x   x2   2x   x3  x  6) f(x)=   x  S:K Lt t i x=5  x  1   x  3) f ( x )   x  S:Lt/ R x   3x  4 x  1     x 4  x  2 S:Lt/ R 4) f ( x )   x   x  2 4 Bài 4: Tìm giá tr c a m đ hàm s sau liên t c t p xác đ nh c a chúng:  x3  x2  x   x2  x    x  S:m=0 x  1) f ( x )   x  S:m=3 3) f ( x )   x 1   x  x  3x  m m   x  x x  x  S: m=2 4) f ( x )   x  mx x 1  x  2) f ( x )  2 S: m=1  mx  x  Baøi 5: Ch ng minh r ng ph ng trình sau ln có nghi m: a) x3 – 2x – = S: f(x) liên t c R f(0).f(3)