Question

设f(x)具有二阶连续导数,f(0)=0,f'(0)=0,f''(0)>0.

在曲线y=f(x)上任意一点(x,f(x))(x不等于0)处做此曲线的切线，交x轴于点(u,0).求lim(x→0)(x*f(u))/(u*f(x))

Answer 1

在曲线y=f(x)上任意一点(x,f(x))(x不等于0)处做此曲线的切线:Y-f(x)=f'(x)(X-x),交x轴于点(u,0),
∴u=x-f(x)/f'(x),
u'=1-[(f'(x)]^2-f(x)f''(x)]/[f'(x)]^2=f(x)f''(x)/[f'(x)]^2,
x→0时u→0-f'(x)/f''(x)→-f'(0)/f''(0)=0,
x*f(u)/[uf(x)]→x/u→1/u'→[f'(x)]^2/[f(x)f''(x)]
→[f(x)/x]^2/[f(x)f''(x)]=f(x)/[f''(x)*x^2]→1/f''(x)*f'(x)/(2x)→1/2.

追问

x*f(u)/[uf(x)]→x/u 这一步不是很明白，f(u)应该是一个变量吧，怎么就这样没有了？