设F(x)连续，且lim x->0 f(x)/x=1,求极限lim[1+∫上线1下线0 f(tx^2)dt]的cotx/ln(1+x)次方怎么算？谢谢

Question

如题所述

Answer 1

y=[1+∫(0,1) f(tx^2)dt]^[cotx/ln(1+x)]
lny=ln[1+∫(0,1)f(tx^2)dt]cotx/ln(1+x)
=ln[1+∫(0,x^2)f(u)du]cotx/[x^2ln(1+x)]
limlny=limln[1+∫(0,x^2)f(u)du]cotx/[x^2ln(1+x)]
=limln[1+∫(0,x^2)f(u)du]cosx/[sinx*x^2ln(1+x)]
=limln[1+∫(0,x^2)f(u)du]/[x^2] (sinx等价x,cosx趋于1，xln(1+x)趋于1）
=lim[f(x^2)2x]/[1+∫(0,x^2)f(u)du]2x
=limf(x^2)/[1+∫(0,x^2)f(u)du]
=0 (limf(x^2)=0)