求极限,高手帮忙

用罗必达法则
lim(x->0)[1-x^2-e^(-x^2)]/[x(sinx)^3]
=lim(x->0)[1-x^2-e^(-x^2)]/x^4
=lim(x->0)[-2x+2xe^(-x^2)]/4x^3
=lim(x->0)[-1+e^(-x^2)]/2x^2
=lim(x->0)[-2xe^(-2x)]/4x
=lim(x->0)[-e^(-2x)]/2
=-1/2

极限为0,对e^(-x^2)泰勒展开3项就可以求了.

不好意思,我看错了,更正一下:

e^(-x^2)=1-1*x^2+1/2*x^4-1/6*x^6+O(x^8)

[1-x^2-e^(-x^2)]/[x(sinx)^3]=

-1/2*[x^4-1/6*x^6+O(x^8)]/[x(sinx)^3]=
-1/2