lim(cosx)^cot^2x-搜答案-搜狗做题网

原式=lim(x->0)e^[cot²xln(cosx)]=e^[lim(x->0)ln(cosx)/tan²x]=e^[lim(x->0)ln(cosx)/x²]=e^

由cot（90°-x）=√2得到tan（x）=√2,所以cotx=√2/2原式分子分母同时除以sinx得到1+cotx/1-cotx=3+2√2

利用对数性质(cosx)^(1/x^2)=e^[ln(cosx)^(1/x^2)]=e^(1/x^2*lncosx)=e^(lncosx/x^2)只要对指数部分求极限即可,有两种方法:一,等价无穷小l

设f(x)=(cosx)^(1/ln(1+x^2)),lnf(x)=ln(cosx)/ln(1+x^2)x→0,ln(cosx)=ln[1+(cosx-1]cosx-1-x^2/2ln(1+x^2)x

洛必达法则上下求导得答案是1/2再答：把cot2x写成cos2x/sin2x

LIM（趋近与0）(cot^2x-1/x^2)=lim(x^2*cos^2x-sin^2x)/(x^2*sin^2x)=lim[(x^2+1)cos^2x-1]/x^4=lim[(1/2)*(x^2+

2/3

sin2x/[(sinx+cosx-1)(sinx+1-cosx)]=sin2x/[(sinx-(1-cosx)(sinx+1-cosx)]=sin2x/[(sin^2x-(1-cosx)^2]=si

1+cosx=1+2(cosx/2)^2-1=2(cosx/2)^21+cosx-------=2(cosx/2)^2/2sin(x/2)*cos(x/2)=cot(x/2)sinx

lim x->pi (x^2-1)/cosx

1-pi*pi(x^2-1)/cosx在点x=pi是连续的,所以代入x=pi就是所求的极限值.

x无限大 lim cosx

-1

(1+cosx)/(1-cosx)+(1-cosx)/(1+cosx)通分=((1+cosx)^2+(1-cosx)^2)/1-cos^2(x)=2*(1+cos^2(x))/sin^2(x)因为1=

求lim(1-cosx)/x^2

答：lim(x→0)(1-cosx)/x²=lim(x→0)2sin²(x/2)/[4*(x/2)²]=lim(t→0)(1/2）(sint/t)²=1/2

lim(1+x²)^cot²x=lim(1+x²)^(1/x²)(x²cot²x)=lime^(x²/tan²x)=e

QQ42779066,和我语聊告诉你

因cosx/2cosx/4…cosx/2^n=[cosx/2*cosx/4*.*2sinx/2^n*cosx/2^n]/(2sinx/2^n)=[cosx/2*cosx/4*...*sinx/2^(n

lim（x→0）=(1/x^2-cot^2x)=lim（x→0）=(1/x^2-1/tan²x)=lim(x->0)(tan²x-x²)/x²tan²

我算了下,你看看行不lim(arcsinx/x)^{[cot(x)]^2}（x→0）=lim[1+（arcsinx-x)/x]^{[cot(x)]^2}（x→0）=lim[1+（arcsinx-x)/

x趋近于0时,tanx→x,cotx→1／x,(1＋x)^(1／x)→e原式＝lim(1＋x)^(1－2／x)＝lim(1＋x)／(1＋x)^(2／x)＝1／e²