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

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.

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 $$g'(0)$$: $$g'(0)=\lim_{h \to 0} \frac{g(h)-g(0)}{h}$$ $$g'(0)=\lim_{h \to 0} \frac{\frac{f''(0)}{2}h+r(h)h}{h}$$ $$g'(0)=\lim_{h \to 0} \frac{f''(0)}{2}+r(h)$$ $$g'(0)=\frac{f''(0)}{2}$$

• Thank you! Is the observation that $g$ is continuous at $0$ especially important (for instance, if $g$ (or any other function) were somehow discontinuous at $0$, could it still be possible in theory to "find" a derivative at $0$ as we did? If so, would this derivative be extraneous?) – Tiwa Aina Nov 14 '18 at 9:53
• $r(x)$ should be $r(h)$ in the computation for $g'(0)$. – egreg Nov 14 '18 at 10:07
• @TiwaAina differentiability requires continuity. If $\lim_{h \to 0} g(x) \neq g(0)$, then the quotient is not in the $\frac{0}{0}$ form, but in the $\frac{something}{0}$ form, which is divergent. – Botond Nov 14 '18 at 10:12
• Thank you @egreg! – Botond Nov 14 '18 at 10:13

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

• @Botond Thanks. That was a typo. Corrected the answer. – Kavi Rama Murthy Nov 14 '18 at 9:52
• Thank you! I thought that I did something wrong. – Botond Nov 14 '18 at 9:54
• @KaviRamaMurthy Would you mind showing the calculations? For some reason L'Hôpital gave me $\frac{f'(x)-f'(0)}{2x}$. – Tiwa Aina Nov 14 '18 at 9:55
• You have got it right up to this stage. Note that you still have $\frac 0 0$ form. If you apply L'Hopital's Rule again you get the answer. @TiwaAina – Kavi Rama Murthy Nov 14 '18 at 9:58
• L'Hospital's Rule can be applied only once in a useful manner here. And then one can use definition of derivative to get $f''(0)/2$. – Paramanand Singh Nov 15 '18 at 14:50

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 $$\mathbb{R}$$.