I am trying to show the following:
Let $I$ be an interval and $f\colon I\to \mathbb R$ be continuous, and differentiable on $I\setminus\{c\}$ with $f'(x)\to L$ as $x\to c$ in $I\setminus\{c\}$. Then $f$ is differentiable at $c$ with $f'(x) = L$.
My attempt at a proof:
Let $\varepsilon > 0$. Since $f'(x)\to L$ as $x\to c$, take $\delta > 0$ such that for each $y\in B_\delta(c)\cap I\setminus\{c\}$, we have $|f'(y) - L| < \varepsilon$. Now, let $x\in B_\delta(x)\cap I\setminus\{c\}$. Now, since $f$ is continuous on $[c, x]$ (or $[x, c]$) and differentiable on $(c, x)$ (or $(x, c)$), by MVT, take a $y$ strictly between $x$ and $c$, and hence in $B_\delta(c)\cap I\setminus\{c\}$, such that $f'(y) = \frac{f(x) - f(c)}{x - c}$. Now, we have that $|f'(y) - L| < \varepsilon$, and hence $\bigl|\frac{f(x) - f(c)}{x - c} - L\bigr| < \varepsilon$. Since $\varepsilon$ was arbitrary, this means that $\frac{f(x) - f(c)}{x - c}\to L$ as $x\to c$ in $I\setminus\{c\}$ as was required.
Looks correct?