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Assume we have a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f$ is differentiable for all non-zero $x$, $f$ is continuous at $0$ and $$\lim_{x \uparrow 0} f'(x) = \lim_{x \downarrow 0} f'(x) < \infty. $$ Is it then true that $f$ is also differentiable at $0$?

If we drop the hypothesis that $f$ is continuous, this is not true since the signums function is a simple counterexample. But what if $f$ is continuous? I don't think this is true but I can't find a counterexample...

math635
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2 Answers2

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Yes, $f$ is differentiable at $0$.

Hint. By the Mean Value Theorem used in the interval $[0,x]$ for all $x>0$, $$\frac{f(x)-f(0)}{x-0}=f'(t_x)$$ for some $t_x\in (0,x).$ Hence $$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^+}f'(t_x)=\lim_{t\to 0^+}f'(t)$$ The same can be done in $[x,0]$ fo $x<0$.

Robert Z
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f is diferenciable at 0. Hint: Use the definition of the derivative at 0, and by continuity of f at 0 numerator converges to 0 and denominator converges to 0 then apply L'hopital's Rule.

DIEGO R.
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