f(x)有二阶连续导数大于0F(0)=F#39;(0)=0u是f(x)在（x,f(x)）处切线在x轴截距,求lim（x→0）xf(u)/uf(x)

问题：f(x)有二阶连续导数大于0F(0)=F#39;(0)=0u是f(x)在（x,f(x)）处切线在x轴截距,求lim（x→0）xf(u)/uf(x)

答案：↓↓↓

彭宏韬的回答：

网友采纳　　由题可知,f(x)=ax²+o(x²)　　u=x-f(x)/f'(x)　　limu/x=lim[1-f(x)/xf'(x)]　　而limf(x)/xf'(x)=limf'(x)/[f'(x)+xf''(x)]=limf''(x)/[f''(x)+f''(x)+xf'''(x)]=f''(0)/(2f''(0)+0)=1/2,所以limu/x=1/2　　且limf(u)/f(x)=lim[au²+o(u²)]/[ax²+o(x²)]=limu²/x²=1/4　　所以原式=[limf(u)/f(x)]/(limu/x)=1/2