3
$\begingroup$

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...

  • 0
    You should also assume that left and right limits of the derivative are finite – 2017-01-07
  • 0
    Right otherwise this is trivial. Thanks – 2017-01-07

2 Answers 2

2

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 $x<0$.

1

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.

  • 0
    as far is I know for lhospital you need that both the denominator and the nominator converge to 0. – 2017-01-07
  • 0
    Yes this is the case – 2017-01-08
  • 0
    i dont think so... the limit only has to be finite. – 2017-01-08
  • 0
    The numerator is f (x) - f (0) , that clearly converges to 0 when x ->0 , the denominator is x - 0, that clearly converges to 0 – 2017-01-08
  • 0
    ah yes you are right – 2017-01-08