Show $f'(0)$ exists and is $L$ Our assumptions for this problem are:
$f$ is continuous on an interval containing $0$ and differentiable for all $x$ not $0$. Moreover, $$\lim_{x \to 0} f'(x) = L.$$
We must show that $f'(0)$ exists and is $L$.
My thoughts so far:
I am able to produce a (somewhat drawn out) argument for the case in which $f'$ is differentiable at $0$ but cannot seem to account for when that is not necessarily the case. Any help in reducing this argument to that end is appreciated.
 A: I'm sure there is a L'Hopital argument, but I don't want to worry about the hypotheses. 
Fix $x > 0$. Since $f$ is continuous on $[0,x]$ and differentiable on $(0,x)$, the mean theorem theorem asserts the existence of $c_x \in (0,x)$ such that
$$\frac{f(x) - f(0)}{x - 0} = f'(c_x).$$
Thus
$$\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} f'(c_x) = L$$
since $c_x \to 0$ as $x \to 0$. Repeat for $x < 0$ for the desired result.
A: L'Hopital's rule applies here quite nicely. Since $f(x)$ is continuous, $\lim_{h\to 0} f(h)=f(0)$, so $$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{f'(h)}{1}=L.$$
A: You must show $\lim_{x \to 0}\dfrac {f(x) - f(0)} {x - 0}$ exists and equals $L$. Use Lagrange's mean value theorem for $f$ on the interval $[0, x]$ (if $x$ is positive, or $[x,0]$ if $x$ is negative). Then use the fact that $f'(\xi)$ has limit $L$ if $\xi \to 0$ with $\xi \not = 0$.
A: I am self-studying Abbott's understanding analysis 2 ed. This problem appears as exercise 5.3.8. I found an alternative proof, and I would appreciate it if you comment on it.
Let $l(x) = f(x)-f(0)$ and $g(x) = x$. By the Generalized Mean Value theorem, there exists a point $c \in (0,x)$ (the case with $x<0$ can be derived analogously) so that $\frac{l'(c)}{g'(c)}=l'(c)=f'(c)=\frac{l(x)-l(0)}{x-0}=\frac{f(x)-f(0)}{x}$.
Taking limits as $x \rightarrow 0$ on both sides of the inequality, and recalling that $0<c<x$ gives us $\lim_{x \rightarrow 0}f'(x)=L=\lim_{x \rightarrow 0}\frac{f(x)-f(0)}{x}$, the right side of the inequality being precisely the definition of the derivative of $f$ at zero, proving the desired result.
