Can a one-sided limit exist at a discontinuity point of a derivative?

Question

Let $f \colon \mathbb{R} \to \mathbb{R}$ be a differentiable function and fix $a \in \mathbb{R}.$ By L'Hôpital's rule, since $$ \frac{(f(x)-f(a))'}{(x-a)'} = f'(x),$$ if either limit $$ \lim_{x \to a^+} f'(x), \qquad \lim_{x \to a^-} f'(x)$$ exists, then it equals $f'(a).$ Hence, if both of them exist, $f'(x) \to f'(a)$ as $x \to a$ and $f'$ is continuous at $a.$ Equivalently, if $f'$ is not continuous at $a,$ then at least one of the one-sided limits does not exist. This is enough to show that every discontinuity of $f'$ is essential, but I'd like to know whether it can be assured that neither one-sided limit exists.

I appreciate any help.

How about $f(x)=0$ for $x\le0$, $ f(x)=x^2\sin(1/x)$ for $ x>0$? — David Mitra, Nov 27 '20 at 15:57
@David, in view of the wrong answer perhaps you should post yours. — Ruy, Nov 27 '20 at 16:01
@DavidMitra Thanks a lot. I see one of the one-sided limits might exist. — sebpar, Nov 27 '20 at 16:15

score 2 · Accepted Answer · answered Nov 27 '20 at 16:16

2

No. Define $f$ by $f(x)=0$ for $ x\le0$ and $f(x)=x^2\sin(1/x)$ for $x>0$. One can show $f$ is differentiable and that $f'$ is not continuous at 0. Obviously, the left-hand limit of $f'$ at 0 exists.

answered Nov 27 '20 at 16:16

David Mitra

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Can a one-sided limit exist at a discontinuity point of a derivative?

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