lim(cosx-2x) sinx x-搜答案-百度作业网

lim((x+sinx)/(x-cosx))

lim(x→0)[(1+xsinx)^1/2-cosx]/sin^2(x/2)=lim(x→0)[(1+xsinx)^1/2-1+1-cosx]/sin^2(x/2)=lim(x→0)[(1+xsin

Taylor展式：原极限＝lim{1+1/2x^2-(1+1/2x^2+(1/2)(-1/2)/2x^4+小o(x^4))}/{[1-x^2/2-(1+x^2)+小o(x^2)]x^2}=lim[1/

当X趋近于0的时候ln(1+x)可以用X代替,所以原式可以变为sinx/[(1+cosx)*x]+x^2*sin(1/x)/[(1+cosx)*x]先求sinx/[(1+cosx)*x]的极限,sin

lim(x→0)[√2-√(1+cosx)]/sin²x,0/0型,洛必达法则=lim(x→0)[sinx/(2√(1+cosx))]/(2sinxcosx)=lim(x→0)1/[4cos

x->π/2吧对分子cosx=sin(π/2-x)因为π/2-x->0所以sin(π/2-x)~(π/2-x)对分母cos(x/2)-sin(x/2)=√2[((√2)/2)cos(x/2)-((√2

tan²-1=sin²x/cos²x-1=(sin²x-cos²x)/cos²x=(sinx+cosx)(sinx-cosx)/cos&su

对分子分母都除以xlim∞>[sin(x^2)–x]/[((cosx)^2)-x]=lim∞>[sin(x^2)/x–1]/[((cosx)^2)/x-1]=lim∞>[sin(x^2)/x–1]/l

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)sin2x5xsin2x2lim-------------------=lim--------------*lim------------*------=2/5x→0sin5xx→0sin5xx

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

limcos（x+π/2)/t=lim-sin^2(t)/t=0这一步错了应该是limcos（x+π/2)/t=lim-sin(t)/t=-1

使用等价无穷小即可求解因为x→0时,e^x-1~x1-cosx~x^2/2所以原式=lim(sinx)^3/(x*x^2/2)=2lim(sinx)^3/x^3又x→0时,sinx~x所以原式=2

先用L'Hospital法则计算lim(x→0)[1/(1-cosx)]*ln(sinx/x)=lim(x→0)ln(sinx/x)/(1-cosx)(0/0)=lim(x→0)[1/(sinx/x)

sin^2x/(sinx-cosx)-(sinx+cosx)/(tan^2x-1)=sin^2x/(sinx-cosx)-(sinx+cosx)/[(tanx+1)(tanx-1)]=sin^2x/(

lim(1-cosx²)/(sin²xtan²x)=lim2sin²(x²/2)/(x²*x²)=2lim(x²/2)&

(1+sinx-cosx)/(1+sinβx-cosβx)=（sinx+2(sin(x/2))^2)/（sinbx+2(sin(bx/2))^2)=[(sin(x/2)+cos(x/2))/(sin(

不对,应该是cos(x/2)lim△x-->0[sin（△x/2)]/(△x/2)=1是等价无穷小.