(78)-高数专项练题1-2_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

2025 专项刷题班第一章 函数、极限、连续x  2tx x 例1.1 设 f (x)  lim ,则F(x)   f (t)dt 在 x  0处( C ). t 1  2tx 1 (A)可导 (B)间断 (C) 连续但不可导 (D)无法判定 YYLX [D] ***, H X10 Xco X =0 X : , # 2X 2 * fix X0Af Etto +0 / = = = , , , ** % ② X10 #J , E+ +c , 2 = z = 0 fix = X , ③ X =0 #J , y + = 20 = 1 f(a 2 = , : fin I 1, X > 0 flot) 1 flo = = , = o E X= 0 , TES = X =0 EDER FIN -I X0 ,↑ DEFNESE Fix ) * findt E E Fix fin = = , . , fin IT E FIN - F SE-ELE(YY(X) , ② , 1 fin 75 B B FIN XotEy GE Xo , EN Mfx N f = IX , = 0 X =0 . Y X+0 & Fix 1 "fitdt FM X = = 1 = = . o , X=0 , , 2 fIN EXo Ye BRID E FIN TXo R]. B .. 1 X 0 (o- fN d . F(x fitdt S Xi X30 * = = = - 1 X0 X . - Xo ,fin E FINE Tz]e 3 Texo Xo , , tr : fix = & , x+o Fix = 1 fint = &kM , x+o . 0 X= 0 0 X =0, . , - FF) Fid Men A = = = co *0 X - 8 X EXo 4 FE FIN Exo JEFFT ,x sin x 2 例1.2 已知函数 f (x)   ecostdt, g(x)   et dt, 则( ). C 0 0 (A) f (x)是奇函数, g(x)是偶函数 (B) f (x)是偶函数, g(x)是奇函数 (C) f (x)与 g(x)均为奇函数 (D) f (x)与 g(x)均为周期函数 fix = etat fast FMBE) , . . et % "et at M . h = , , 1 . ** et at hismx) B. Ex 9ix = = . est? T 127A, post 2 fix F = at >0 , (2↑ 例1.3 当 x  0时，下列无穷小中，阶数最高的是( C ) (A)ln(1 x2)  x2 (B) 1  x2  cos x  2 2 x 2 (C) ln(1  t2 )dt (D)ex  1  x2 0 In CHE) -** 1 X 4 (x - (n(1 + x) 2x) (A) ~-- = Y Ex (B) It* + 103X -2 = It - +ON) + (1 - 2 x + X + OmY) - 2 Ex* = o() Ex - + v - * ()( ti /x InG + at Edt ↓ 5X = = Xy (p)eX x 1 - - - z1  x 2arctan x  ln 1  x 例1.4 已知lim  c  0,则( A ) x p x0 4 4 4 4 (A) p  3,c   (B) p  3,c  (C) p  ,c  3 (D) p   ,c  3 3 3 3 3 InCX) In -x ZarceX- + 7 /m = XP - EX EXP X o() (x TONY1 ((X) (3) u 2(X + = - - + - To #0 XP ** 43 0( P 3 mm - + M - 5 : = = = C + 0 * ~ Xi = ↑ XTj XP C - 5例1.5 设 y  y(x)是方程 y  2 y  y  e3x 的解，且满足 y(0)  0, y(0)  0，则当 x  0时，与 y(x)为等价无穷小的是( D ). EX (A)sin x2 (B)sin x (C)ln(1 x2) (D)ln 1  x2 ~ X 2 NX ~ X ~ 1.) + GXeY r n Ge 20 1 => = = - = + + = 0 (5 = - * + y AeY Ax y" y y A = + + = e = = , => C C . . . , y" 24 y() 33 = = (10) = 0, %10 =0 #x + +y + e - * = = 1 2 + o Mm 40 70X = +o = +     例1.6 已知数列 a a  0 , 若 a 发散, 则（ D ）. n n n .  1   1  X(A) a  发散 (XB) a  发散 n n a a     n n  1   1  X a a (C) e n  发散 (D) e n  发散 a a e e  n   n  = Man 2455 FREE [an RD , n [T & (B) (C) (2/31 an = . . F #Ran 92 n = = , & (A) R(3 , = n  1  t2 sin t2 x 例1.7 当 x  0 时, dt 与 xk 是同阶无穷小, 则 0 1  cost2 k  ____3___. I 2 ↑ anot El + 2 AJ -: to t , ~ z 2 Itlost ~ > 2 I↓ C + E) ·Sit l Ent n at + -t = X3 = = ~ lost It↑  3 1  例1.8 lim x sinln(1  )  sinln(1  ) ________.   x x x   ESTHE H * FREHA 2107 E 35- FE : , , SmI(r) 5 + * cosM5 ) - smmG = . , (SIulux'Is Elvis cos1uX . ME = . = los S = * REMS , = -2 3 1  例1.8 lim x sinln(1  )  sinln(1  ) ________.   x x x   XSMICHXSm(n(+* 35 75 ) == = InC)-Wx (c+ ) In t . = X . . UX MX = 3 1 = - = 22 (1  x)x  e2[1  ln(1  x)] 例1.9 求极限lim x x0 Melex) -e Elle + = X In CTX) 2 E Inc ) may +N Ie ed - = t * X X 2 InC X) + 2 - Me(e 1) 2 = - e + ↓ = ? ( InCX) 2) im e - 2 = C + #0 Y= ? ( InCX) 2) im e - 2 = C + #0 Y 2x) (2mCTX) ume - 2 = 2 * # X (- ? ze m = C + X X =O 1 cos2 x  例1.10 lim    sin2 x x2 x0   sinex - & X -suix cos X Im - Te n - = 2 Shix xo XY & -M2x- 1. DSM2X 10S2X & . . M2X-ESU 35 = 4X 3 4X (4x1 4X-SmyX Im Im = 64 = 4 = #o 8X *b 8x3 = 5 48 IX-Esm2x (x I Smix) , 15 M + . =: = /m (2x + Sm2x) (2X-SMIX) = X4 # ext (x Sm2X (2x M + I Im 2x+Sm2X 4 = = 5 = 5 ↑ X)0 . 4 X! 1  cos x cos2x n cosnx 例1.11 求极限lim ( x2 x0 "con G m1-elucox cosx ... TE 35- : = X +... cosx Icony M1-eIncox + = 2 X Incosux) (IncosX &Incos2X m + +...+ = - #O Xa 33 - SmX 1-2SM2X I as my ( M (cost + & .. + I z = - CoSIX #0 2x↳ Smx1-2Sm2X -usmy - a ( M ( & & + I =CoS2X - losX #0 2x X pa tax + tan2X +... + tak = #O 2X ( )nn( ! ((+ 2 + + ) + = = 2 -1  cos x cos2x n cosnx 例1.11 求极限lim x2 x0 " 35 T f 1-coSX + losX-losN cosX ... con =: = X Men 1-103X MycosX(1-cosex " cosuy = ... + ? Ar X X cosX-cosx I 11-cX + = I + X2 1-10sx Me (1- " co I M cosx. = + + . X2 Ap X " cou Mm(1-cosx I ... = 1 # = + + I X2 n(utt) = 4x  (3  2tan t)t  3t dt   例1.12 计算极限lim 0 . 3 x0 e3x  1 # 2 tart 3t) [(3 at + - 73 I = o 3x mx)" 34 X In (3 +2 tmX) Xlu3 25 (3 + 2+ - Im e C - = 9x o Xto 9x2 ↓ In[3 +2TmX) XIn3 exe - C I - I = 9x2 = um X In (3 +2taX) - Xlu3 Jum In (3 + 2tax) - In 3 - X 9x2 #0 9XIn In (3 + 2tax) In3 - = #0 9X tmx 2 / m(l Im + = * 9x 2EY Im = 3 ar 9X 2 =