题目

lim(x->0) 1-x^2-e^(-x^2)/x*sin^(3)2x

提问时间：2020-12-03

答案

∵lim(x->0){[1-x²-e^(-x²)]/(8x^4)}
=lim(x->0){[1-x²-e^(-x²)]'/(8x^4)'} (0/0型极限,应用罗比达法则)
=lim(x->0){[e^(-x²)-1]/(16x²)} (求导化简)
=lim(x->0){[e^(-x²)-1]'/(16x²)'} (0/0型极限,应用罗比达法则)
=lim(x->0){[-e^(-x²)]/16} (求导化简)
=-1/16；
又lim(x->0){[(2x)/sin(2x)]³}
={lim(x->0)[(2x)/sin(2x)]}³
=1³ (应用重要极限lim(t->0)(sint/t)=1)
=1；
∴lim(x->0){[1-x²-e^(-x²)]/[x*sin³(2x)]}
=lim(x->0){[(2x)/sin(2x)]³*[(1-x²-e^(-x²))/(8x^4)]}
=lim(x->0){[(2x)/sin(2x)]³}*lim(x->0){[1-x²-e^(-x²)]/(8x^4)}
=1*(-1/16)
=-1/16.