TITLE: Derivative has finite, unequal left and right limits at a point; is the function non-differentiable at this point?
QUESTION [5 upvotes]: I have a short question, related to the ongoing search of mathematics instructors for counter-examples to common undergraduate mistakes.
The classical example of a function that is differentiable everywhere but has discontinuous derivative is
\begin{equation}
   f(x)=\left\{
      \begin{array}{cc}
         x^2\sin(1/x) &(x\neq0), \\
         0            &(x=0),
      \end{array}\right.
\end{equation}
which has derivative
\begin{equation}
   f'(x)=\left\{
      \begin{array}{cc}
         2x\sin(1/x)-\cos(1/x) &(x\neq0), \\
         0            &(x=0).
      \end{array}\right.
\end{equation}
$f'$ fails to be continuous at $0$ purely because its left- and right-hand limits do not even exist at $0$.
However, suppose that we have found a function $g$ whose derivative $g'$ has finite but unequal left- and right-hand limits at some cluster point $x_0$ in its domain. May we conclude that $g$ is not differentiable at $x_0$?
If this is not the case, is there a simple counter-example? (I'm guessing such a counter-example ought to be more complicated than the $f$ I have given above, as $f$ is sometimes claimed to be the simplest example of a differentiable function with discontinuous derivative.)
Thanks in advance!

REPLY [2 votes]: Suppose $f$ is differentiable over $(x_0-k,x_0+k)$, for some $k>0$, and that
$$
\lim_{x\to x_0^-}f'(x_0)=l
\qquad\text{and}\qquad
\lim_{x\to x_0^+}f'(x_0)=r
$$
with $l\ne r$. It is not restrictive to assume $l<r$ (otherwise take $-f$).
Take $\varepsilon=(r-l)/4$. Then there exists $\delta$ with $0<\delta<k$ such that
\begin{align}
f'(x)&<l+\varepsilon &&\text{for every $x\in(x_0-\delta,x_0)$}\\
f'(x)&>r-\varepsilon &&\text{for every $x\in(x_0,x_0+\delta)$}
\end{align}
This contradicts Darboux’s theorem, because $f'(x)$ should assume every value in the interval $(f'(x_0-\delta/2),f'(x_0+\delta/2))$, for some $x\in(x_0-\delta/2,x_2+\delta/2)$. But we see that infinitely many values are missed, as $f'(x_0)$ can only provide one of them.