(79)-高数专项练题2_08.2026考研数学高途王喆全程班_赠送2025课程_25考研数学（三）全年智达班_{2}--资料

(1  sin x )(1  3 sin x ) (1  n sin x ) 例1.13 lim  (1  sin x)n1 x 2 : / = 3 - t t -X X = , (1-sm) (l-smt) (1- "SuCE T . ... Um = + [1-SuCE-t)" (1-lost) (1 -Most (1- "cost) · ... In = To + Cl-cost" I I I Z 2 5 t in th t um Z - .... . I = 2 + 2 [xyx x Ito = ... = N ni ⑳th gu+1  f (x) x f (x) 例1.14 已知lim 1  x   e3,lim   x x2 x0   x0 f() In( : +x + e e ↓ = to + In Cit M => 3 = * X f) In 1 + x+ FigABX 35- : X+O #J 3+ d(x + 0) , = , Y + x(3 d) => In (1 + x ) = x (3 + 2) = H+ x+ + e = (3+ ex => 1 X = - -m M-X eXB => M = = / - M X = = + 35 Mum(+ = : = AMX 3 = = * X In Ux = M => xt Im => I = 3 = * Mo = 2 =1  x 例1.15 设函数 f (x)  lim , 则 f (x)（ ）. 1  nx2n n (A) 在 x  1, x  1处都连续 (B) 在 x  1 D 处连续, x  1处不连续 (C) 在 x  1, x  1处都不连续 (D) 在 x  1处不连续, x  1处连续 X (x* "She * C X31 8 [ = = = , . = # DECIAF Y x : new -0 n - 0 , , , . fix : 1+X = Y (X fix ② EX2 1 At nto n = n . + 0 0 = , , , . " fill Ms it ③ EX = 1 #J n = n , = = O , fil EX ④ = -AJ = o ,X i 1xx f(x ((+x. = : = X1X : -1 > 0 . f(l 0 fit = ME (X) = 0 fH) = 0 : + * = . , f f 0 (i M 2 = = =例1.16 设函数 f (x)在 x  0处可导，且 f (0)  1, f (0)  3,求数列极限 1  1  1 n(1cos ) I  lim f ( ) .   n n n  I fitl In fast M not 1 & = = fint-1] flu)-1 In [l + m me = e n.til = n+ fin)-fiol eafin 2. 6 I C = e = I = - 0 i2 2 2  1   2   n  例1.17 limln n 1  1  1  等于( B ).       n n n n       2 2 2 2 (A) ln2 xdx (B)2 ln xdx (C)2 ln(1  x)dx (D) ln2(1  x)dx 1 1 1 1 m(h] (H) |(H+E 7. Uns[ In + + ... + = In [mC 2M + = =I m InCl x] + dx = 2 t = HX 2/2 me /2 mx de 2 dx = I例1.18 设 x  0, x  1  e x n ,(n  1,2, ).(1)证明lim x 存在，并求此极 1 n1 n n 限； SERA X 7 0 (1) : , , XI - 1-e < : 07Xz = * : xX 1 e = - i i 17 SIA 35 It o < Xn < /1-EM 35 Xnt-Xn Xn - : = - * 1 (0(X 1) 9m 1 -e - x - 91 % = 0 = , * gin - 1 = e - -X0 gix) 9 V 0 : < , 9I 910) : < = 0 Xnt1 < Xn [x] NATO :e-X fix 1 - eX /2 1 (0xx() 3 = = Xn = - , = ex fix > 0 = * 1137 A P [x] : . . P :H3) X a , Ve M(-eX Xnt , 1 = nico a - 1 e a - = : a = 0 nu Xn 0 im = Hesx  x (2)求lim n1 n . x 2 n n - Ml-e -x et An Xnt-X2 Im 1 t - - = Yu Xu nico to0 th 1) (m (e - (-4) - Etti - My- = - = Eto -2 to = 2 -    例1.19 设 f x 是区间 0,  上单调减少且非负的连续函数， n n a   f  k    f  x  dx  n  1,2,  ，证明数列  a  的极限存在. n n 1 iK1 fin)-S "fix fill fill G EDA : an = + +...+ ax , fill fill fini finti) 14t fixdx Any = + +... + + - " 1 finti fixax 9" fix Any-an = - + ex : , , firs (frfixax-f"f(xax) = - . fintil In fix ax = - fint1)-flul EFEY SuE(n (i5 : IRIE Gut-C ntl) = . , and fix V . Ant-an : 10. -C * (*1ax-fr* (35 Ant-an fint fixax final fixdx == = - = Crfint-fix] Gut An) > 0: anTTB0 : M CTP - .x4 1 例1.20 曲线 y  arctan 渐近线的条数为( B ). x2  1 x . (A) 2 (B) 3 (C) 4 (D) 5 T M : DKE Mr aritmt : · = = 0 , /nTd7- : EX ② E X = - 1 X= 1 x0 : , , . xY I m ariem - · +)X 1 x - TEE- X + 5 X =1 : = -⑪t o 1 ↳ m an = 0 + 0... X =0 RAY FRE ↓ # - 14 ↑ ③ x3 avcint = I Am = 1 = - # manm-x] b - Ef = acctit - to Ez-1 concent My concert-tc-ti E m) E] : : Y X = = = - to t +Y E0 (1 t)) - t - t -to(E) Pe constant + + A t - Mr = - + = #or 2 = O + e2ln x 例1.21 设函数 f (x)  sin x，则 f (x)有( ). A x  1 (A) 1 个可去间断点，1 个跳跃间断点 (B) 1 个跳跃间断点，1 个无穷间断点 (C) 2 个可去间断点 (D)2 个无穷间断点 EX X= 15X =0 . lux Ink Un It jun smx sml SmM = = x1(x 11 -1x 1) - - + sml/m x f(i) filt sml = = - sm/ , = x (x 1) - 1 -** :Ink (n(x jun smx MX =0 = . & 01x-11 + * 33 X= 0 - --1 1 1 1  例1.22 设 f (x)    , x  ,1 , 试补充定义使得   x sinx (1  x) 2    1  f (x)在 ,1 上连续.   =  2  P fill I I fit = J + T M - : smax x( x) - I r(smax ( =+ - acces I t 1-X smae = m] 1) sit) + - = + - tot Smirk-t) Xt + -(x Smit 7t- pr + mm = = + = + tot 7t Smit z to thax  ln(1  x) 例1.23 设函数 f (x)  的可去间断点为 x  0，则a,b满足 x  bsin x ( ). (A)a  1,b 任意数 (B)a  1,b 任意数 (C)a任意数 b  1 (D) a D 任意数 b  1 P 05 # x= , ax-InCITN) nu * E = X+ o Afax-InCI + X) Bil > X + bsmX . bSmX # ** o() 2 (x E => Max - - + im Ga-l . X + + oix = x0x o(xi) b(X + x *0 + - + (1 + b) x -X o() - + a b 1 : + ,