Is $f$ differentiable at $0$? Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function which is differentiable on all of $\mathbb{R}_{0}$ and suppose that $\lim_{x \rightarrow 0}f'(x)$ exists and is finite. I was wondering if you could then conclude that $f$ is differentiable on $0$ ?
 A: L'Hopital's Rule states that if $\lim_{x\to0}F(x)=\lim_{x\to0}G(x)=0$ then
$$\lim_{x\to0}{F(x)\over G(x)}=\lim_{x\to0}{F'(x)\over G'(x)}$$
provided the latter limit exists.  Let $F(x)=f(x)-f(0)$ and $G(x)=x$, for which
$$\lim_{x\to0}{F'(x)\over G'(x)}=\lim_{x\to0}{f'(x)\over1}=\lim_{x\to0}f'(x)$$
exists (by hypothesis). Then
$$f'(0)=\lim_{x\to0}{f(x)-f(0)\over x-0}=\lim_{x\to0}{F(x)\over G(x)}=\lim_{x\to0}{F'(x)\over G'(x)}=\lim_{x\to0}f'(x)$$
A: Assume that $\lim_{x \to 0} f'(x) = L$.  
Fix $\epsilon > 0$. Choose $\delta$ so that $0 < |x| < \delta \implies |f'(x) - L| < \epsilon.$
Suppose $0 < |x| < \delta$. The mean value theorem gives you a point $y$ in between $0$ and $x$ with $\dfrac{f(x) - f(0)}{x-0} = f'(y)$. In particular $0 < |y| < \delta$ too, so that $$ \left| \dfrac{f(x) - f(0)}{x-0} - L \right| = |f'(y) - L| < \epsilon.$$ Thus $$0 < |x| < \delta \implies \left| \dfrac{f(x) - f(0)}{x-0} - L \right| < \epsilon$$ so that $f'(0) = L$ by the definition of the derivative.
A: $f$ is differentiable at $0$ if
$$
\lim_{\substack{x\to 0\\x\neq 0}}\frac{f(x)-f(0)}{x-0}
$$
exists. Using the Mean Value Theorem you get $\xi_x\neq 0$ with $|\xi_x|<|x|$ such that $f(x)-f(0)=xf'(\xi_x)$. This yields
$$
\lim_{\substack{x\to 0\\x\neq 0}}\frac{f(x)-f(0)}{x-0}=\lim_{\substack{x\to 0\\x\neq 0}}f'(\xi_x)=\lim_{\substack{y\to 0\\y\neq 0}}f'(y).
$$
In the last step I substituded $y:=\xi_x\to 0$ for $x\to 0$ since $|\xi_x|<|x|$.
Therefore the differential quotient exists and $f$ is differntiable at $0$.
