Question

求极限 limx(sin1/x^2)^1/2 x→－∞

Answer 1

limx(sin1/x^2)^1/2 x→－∞

=lim[-(-x)](sin1/x^2)^1/2

=-lim[(-x^2)sin(1/x^2)]^1/2

=-lim[x^2sin(1/x^2)]^1/2

=-lim{sin(1/x^2)/[1/x^2]}^1/2

当x→－∞时 1/x^2→0

令t=1/x^2,代入

原式=-lim(sint/t)^1/2

又t→0 lim(sint/t)=1,所以

原式=-1

一般情况下,这类题都要转化成lim(sint/t)的形式哦~