Showing differentiability of $g(x)=\begin{cases}\frac{f(x)}{x},&\text{$x\neq0$}\\f'(0),&x=0\end{cases}$ given that $f(0)=0$

Question

Let $f$ be a twice-differentiable function of $\mathbb R$ with $f(0)=0$. Define $$g(x)=\begin{cases}\frac{f(x)}{x},&\text{$x\neq0$}\\f'(0),&x=0\end{cases}$$ Prove that $g$ is a differentiable function of $x \in \mathbb R$.

I tried using the difference quotient around $0$ to get

$$g'(x)=\lim_{\epsilon \rightarrow 0} \frac{g(\epsilon)-g(0)}{\epsilon}=\lim_{\epsilon \rightarrow 0} \left( \frac{f(\epsilon)}{\epsilon^2} - \frac{f'(0)}{\epsilon}\right)$$ but this doesn't seem to be of much use. Apparently the problem can be solved using Taylor series, but I fail to see how.

Answer 1

By Taylor's theorem: $$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2}+r(x)x^2$$ $$f(x)=f'(0)x+\frac{f''(0)}{2}x^2+r(x)x^2$$ So $$\frac{f(x)}{x}=f'(0)+\frac{f''(0)}{2}x+r(x)x$$ So $g$ is continuous at $0$. Let's calculate $g'(0)$: $$g'(0)=\lim_{h \to 0} \frac{g(h)-g(0)}{h}$$ $$g'(0)=\lim_{h \to 0} \frac{\frac{f''(0)}{2}h+r(h)h}{h}$$ $$g'(0)=\lim_{h \to 0} \frac{f''(0)}{2}+r(h)$$ $$g'(0)=\frac{f''(0)}{2}$$

Answer 2

Use the Taylor development at $0$, $f(x)=xf'(0)+x\cdot \mathcal{O}(x)$ implies that

$\displaystyle{{g(x)-g(0)}\over{x}}={{f'(0)+\mathcal{O}(x)-f'(0)}\over x}$ where $\displaystyle\lim_{x\rightarrow 0}\mathcal{O}(x)=0$, this implies that $g$ is differentiable at $0$ and its differentiable on $\mathbb{R}$.

Answer 3

$\lim_{x\to 0} \frac {f(x) -xf'(0)} {x^{2}} =\frac {f''(0)} 2$ by L'Hopital's Rule.

Answer 4

First notice that $g$ is continuous at $0,$ by definition of $f'(0):=\lim_{x\to0}\frac{f(x)-f(0)}x.$

Moreover, as $x\to0,$ $$\begin{align}g'(x)&=\frac{xf'(x)-f(x)}{x^2}\\&=\frac{x\Big(f'(0)+xf''(0)+o(x)\Big)-\left(0+xf'(0)+\frac{x^2}2f''(0)+o(x^2)\right)}{x^2}\\&\to\frac{f''(0)}2. \end{align}$$

Therefore, $g$ is continuously differentiable at $0$ and $g'(0)=\frac{f''(0)}2.$