Let $f\colon \mathbb{R}\rightarrow\mathbb{R}$ be a continuous function on $\mathbb{R}$. Suppose $f(x)$ exists for all $x\neq0$, and $\lim_{x\rightarrow0}f'(x)$ exists. Show that $f'(0)$ exists.

I totally have no idea how to prove this, if by definition, we should start with $$f'(0)=\lim_{x\rightarrow0}\frac{f(x)-f(0)}{x-0}$$ and show that this limit exists. But how can I evaluate the limit since the function is given in general.

  • $\begingroup$ To show $f'(0)=\lim_{x\rightarrow0}f'(x)$ . $\endgroup$ – Syuizen Jan 25 '15 at 7:06
  • $\begingroup$ are you sure you mean to say $f(x) $ exists or $f^{'}(x)$ exists $\endgroup$ – Learnmore Jan 25 '15 at 8:10

I'll show that both Left hand derivative and Right hand derivative exists and are both equal i.e. $$\lim_{h\to 0^+}\frac{f(h)-f(0)}{h}$$ and $$\lim_{h\to 0^-}\frac{f(h)-f(0)}{h}$$ exists and are equal.By L'Hopital's Rule $$\lim_{h\to 0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^+}\frac{f'(h)}{1}=\lim_{h\to 0}f'(h)$$ The other limit can be evaluated similarly it will be same. Thus $f'(0)$ exists.


Let $b = \lim_{x \to 0} f'(x)$. After replacing $f(x)$ with $f(x) - bx$ if necessary, we may assume $b = 0$.

Let $\varepsilon > 0$ be given. Then there is some $\delta > 0$ such that $|f'(x)| \leq \varepsilon$ for $x \in (0,\delta]$. It follows from the mean value theorem that for all $x \in [0,\delta]$, we have $$|f(x) - f(0)| \leq \varepsilon x.$$ Thus $0<x<\delta$ implies $\left|\frac{f(x) - f(0)}{x}\right| \leq \varepsilon$, proving that $f_{+}'(0) = \lim_{x \to 0, x > 0} (f(x) - f(0))/x = 0$. The relation $f_{-}'(0) = 0$ can be proved similarly.

Note. The version of the mean value theorem used here is as follows. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with $|f'(x)|\leq M$, then $|f(b) - f(a)| \leq M(b-a)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.