calculus - Derivative has finite, unequal left and right limits at a point; is the function non-differentiable at this point?

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!