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I already know that if a continuous function $f$ is left differentiable and right differentiable in $x$ and the value of those derivatives is the same, than $f$ is differentiable in $x$ and $f'(x)=f'_+(x)=f'_-(x)$. I'm wondering if I'm allowed to say the same if $f\in\mathscr{C}^1$ on both $(x-\varepsilon,x)$ and $(x,x+\varepsilon)$ and \begin{equation} \lim_{y\to x-} f'_-(y)=\lim_{y\to x+} f'_+(y)\ne \infty. \end{equation}

Riccardo Galbusera
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  • since the both derivative are finite and equal that means function is differentiable – mathophile Sep 07 '21 at 16:01
  • Try applying L'Hopital's rule to compute $\lim_{y \to x^-} (f(y) - f(x))/(y - x)$ and $\lim_{y \to x^+} (f(y) - f(x))/(y - x)$. – Dan Velleman Sep 07 '21 at 16:11
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    Your weaker conditions say nothing about the value of $f$ at $x$. It might not even be continuous there. – Karl Sep 07 '21 at 17:24
  • Yes, we need to know that $f$ is continuous at $x$. But if it is, then you can apply L'Hopital's rule, as suggested in my previous comment. – Dan Velleman Sep 07 '21 at 18:05
  • See here: https://math.stackexchange.com/questions/257907/prove-that-fa-lim-x-rightarrow-afx – Hans Lundmark Sep 07 '21 at 19:05

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The function $f = \mathbf 1_{[0,\infty)}$, i.e. the function with value $1$ in $[0,\infty)$ and $0$ in $(-\infty,0)$ is a counterexample. Indeed, $$ f∈C^1((0-\varepsilon,0)\cap(0,0+\varepsilon)) \\ \lim_{y\to 0-} f'_-(y) =\lim_{y\to 0+} f'_+(y) = 0 $$ But $f$ is not continuous at $0$.

LL 3.14
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