Derivative of continuous function exists if limit of derivative exists I'm stuck on this old qualifier problem.  I suppose one could do it using the basic definitions of continuity and differentiability, but is there a simpler way? (For example, using DCT, FTC, Lebesgue differentiation theorem, etc.)
Let $f:\mathbb{R} \mapsto \mathbb{R}$ be continuous.  Suppose $f$ is differentiable away from $0$ and lim$_{x \to 0} f^\prime(x)$ exists.  Show $f^\prime(0)$ exists. 
 A: By the mean value theorem, there is a $c_x\in (0,x)$ resp. $c_x\in (x,0)$, depending on whether $x > 0$ or $x < 0$, such that
$$\frac{f(x)-f(0)}{x} = f'(c_x).$$
As $x\to 0$, by the squeeze lemma, also $c_x\to 0$, hence
$$\lim_{x\to 0} \frac{f(x)-f(0)}{x} = \lim_{x\to 0}f'(c_x)$$
exists.
A: By L'Hôpital's rule, we know that for any function $f$, defined in a neighbourhood of $0$,
$$
\lim_{x\to0}\frac{f(x)-f(0)}{x}=\lim_{x\to0}f'(x)
$$
provided that $f'$ is defined in a punctured neighbourhood of $0$, and the limit on the RHS exists. Since we are given these hypotheses in the question, it follows that $f'$ is both defined and continuous at $0$.
Although this is not directly relevant to the question at hand, it's worth noting that it is possible for $f'(0)$ to exist even if $\lim_{x\to 0}f'(x)$ does not. Consider, for instance, the function
$$
f(x)=\begin{cases}
x^2\sin(1/x) & x\neq0 \, , \\
0 & x = 0 \, . 
\end{cases}
$$
Note also that this answer is essentially equivalent to Daniel Fischer's, as L'Hôpital's rule is derived from the mean value theorem.
