极限计算15天打卡营题目

1、sin xsin x11.求极限1im2x01n(1x1【斛】1im2x01n(1x)22sinx1n(1x)ln(122x)sinx.2sinx1imx01n(1x.r1im)等价代换）.2sin x1im 4x 0x4x2)x21n(14x(sinx1imx0x)(sinxx)(2x)(4x13、6x)lim14x2_4x2.求极限11imx01n(1x2)2、1n(1tanx),，2、1n(1 tan x)1【斛】1im2x01n(1x)1m221n(1tanx)1n(1x)ln(1x2)1n(1tan2x)1im蛇x022tanx)1n(1x)（等价代换）tan2xx2（拉格朗日中值定

2、理）limx0(tanxx)(tanxx)1 3lxmo(2x)x123)33.求极限limtannn【解】limntannlimn1tan()14n又limntan(411)nlimn,，1、,tan(-)tan4n412seclim n（拉格朗日中值定理）故原式4.求极限x(e2xe【解】原式lim2sec一4nx、e)x2xeenxex2xnxee.enlimx05.求极限limx(x1)(x2)(xn)【解】原式limxlimxn(n1)-2-.16.求极限limnnlimnnlimn1nln(1)n7.设alimnlimnlimnlimn0,a1nln(1)nnln(1ln(11)n

3、-)n11,且limxp(axx1a-1)x(e(xlna,则1x)12ln(1x)-x)2六)（拉格朗日中值定理）xpa.1lna(-xInalimxx(x1)Inalimxx(x 1)xp|x|sin(x2)8.函数f(x)-q在下列哪个区间内有界()x(x1)(x2)(A)(1,0)(B)(0,1)(C)(1,2)(D)(2,3)【解1】选(A)排除法因为limf(x)lim|x|sin(x2x1x1x(x1)(x2)2一、|x|sin(x2)|x(x2)limf(x)lim-!2lim2x2x2x(x1)(x2)x2x(x1)(x2)所以f(x)在(0,1),(1,2),(2,3)内均

4、无界,故应选(A).【解2】排除法|x|sin(x2)由于f(x)F在区间(1,0)上连续，且极限x(x1)(x2)|x|sin(x2)limf(x)lim-!2x(1)x1x(x1)(x2)|x|sin(x2)(x)sin(x2)limf(x)lim-!；lim；x0x0x(x1)(x2)2x0x(x1)(x2)2都存在，则f(x)在区间(1,0)上有界，故应选(A).【注】解2中用到一个基本结论：若f(x)在开区间(a,b)上连续，且f(a)和f(b)都存在，则f(x)在开区间(a,b)上有界.9.求极限lim%材x1/1.xx2sinx【解】分子和分母都提因子X得,、=11/1,(x),

5、421原式limxx-X(MJSin2xxlimx1x10.limJ12.n<12.(n1)n-【解】有理化得原式lim,nn.12.n,12.(n1)lim nnn(n1)n(n1),2,2【注】本题用到求和公式12nn(n1).211.若limx03Sinx4-xf(x)3x2,则当x0时，f(x)是x的(A)等价无穷小(B)同阶但非等价无穷小(D)低阶无穷小(C)高阶无穷小【解】一一3sinx由lim4x0x4f(x)3x2可知limsinx4"x3f(x)一3xlimx 0f(x)2x3sinx3-xlxm0则当x0时，f（x）是x的高阶无穷小.12.求极限lim4xl

6、imx3x2Vx42x3)x【解】lim (3. x3x3x2 vx4 2x3) lim x(313 x（等价代换）13.求极限limx 0.1 tanx 1 sinxx(1 cosx)limx(3：131)(4：11)xx，x1312ximx(3?(4(?).1 tanx 1 sinx limx 0x(1 cosx)1r.，tanxsinx2jlimx0x(1cosx)1tanx1cosxlim2x0x(1cosx)1tanx1lim2x0x22xe14.已知limx 0cos2xbax1,求a,b的值.1lim0cos2x bax2x.e lim -x 01 cos 2x 1baxx2 l

7、imx 0122(2x)2b ax（等价无穷小代换）3x2妈获则a3,b2.15.函数 f (x)xx 1x(x 1)ln x的可去间断点的个数为(A)0.(B)1.(C)2.【解】应选C.xx1f(x)在x1,0,1处没定义,x(x1)lnxlim f (x)x 1xim1xx 1x(x 1)ln xlimxxln|x1 x(x 1)ln xx(x 1)ln xlim 一x 1 xlim f (x)x 0lxm0Ix x 1x(x 1) ln x1mxlnlx/e1x(x 1) ln xM0xln xx(x 1) ln xlim 1x 0x 1lim f (x)x 1limx 1|xx 1x(x 1) ln xlimx 1xln|xx