Suppose $f(x)$ has continuous second-order derivative and satisfy $f(0)=0,f'(0)=0,f''(x)>0$. The tangent at the point $(x,f(x))(x \neq 0)$ of the curve $y=f(x)$ intersects the $x$-axis at $(u,0).$ Find $\lim\limits_{x \to 0}\dfrac{xf(u)}{uf(x)}$.
Solution
Expand $f(x)$ at $x=0$ according to Taylor's formula，we have $$f(x)=f(0)+\frac{f'(0)}{1!}(x-0)+\frac{f''(\xi_x)}{2!}(x-0)^2=\frac{f''(\xi_x)}{2}x^2,\tag{1}$$where $0 \gtrless \xi_x \gtrless x$. Moreover，it's simple to obtain the tangent equation as follows $$Y-f(x)=f'(x)(X-x).$$Let $Y=0$. Then $$u=X=\frac{xf'(x)-f(x)}{f'(x)}.\tag{2}$$As per $(1)$，we have$$\frac{f(u)}{f(x)}=\frac{1/2\cdot f''(\xi_u)u^2}{1/2\cdot f''(\xi_x)x^2}=\frac{f''(\xi_u)u^2}{ f''(\xi_x)x^2},$$As per $(2)$，we have $$\frac{u}{x}=\frac{xf'(x)-f(x)}{xf'(x)}.$$Notice that，when $x \to 0$，$u \to 0$（which can be proven using $(2)$ and L'Hospital's rule ），and $\xi_x,\xi_u \to 0$，and consider the continuity of $f''(x)$ and $f''(x)>0$. It follows that$$\begin{aligned}\lim_{x \to 0}\frac{xf(u)}{uf(x)}&=\lim_{x \to 0}\frac{xf''(\xi_u)u^2}{ uf''(\xi_x)x^2}\\&=\frac{f''(0)}{ f''(0)}\lim_{x \to 0}\frac{u}{x}\\&=\lim_{x \to 0}\frac{u}{ x}\\&=\lim_{x \to 0}\frac{xf'(x)-f(x)}{xf'(x)}\\&=\lim_{x \to 0}\frac{f'(x)+xf''(x)-f'(x)}{f'(x)+xf''(x)}\\&=\lim_{x \to 0}\frac{xf''(x)}{f'(x)+xf''(x)}\\&=\lim_{x \to 0}\frac{f''(x)}{\dfrac{f'(x)}{x}+f''(x)}\\&=\lim_{x \to 0}\frac{f''(x)}{\dfrac{f'(x)-f(0)}{x-0}+f''(x)}\\&=\frac{f''(0)}{f''(0)+f''(0)}\\&=\frac{1}{2}.\end{aligned}$$