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

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