Q:

$\lim_{n\to\infty} f_{n}(x) = g(x) \implies \lim_{n\to\infty} f_{n}^{'}(x) = g^{'}(x) $

Let $f_{n} :\mathbb R \to\mathbb R$ be differentiable for each $n \in\mathbb N$ with $|f_{n}^{'}(x)|  ≤ 1$ for all $n$ and $x$.
Assume $\lim_{n\to\infty} f_{n}(x) = g(x) $ for all $x$. Is $\lim_{n\to\infty} f_{n}^{'}(x) = g^{'}(x) $?
I believe it is true, and that one can show it by taking the appropriate limits with respect to some $h \to 0$.
Something of the sort:
$\lim_{n\to\infty} f_{n}(x) = g(x)$
$-\lim_{n\to\infty} f_{n}(x) = -g(x)$
$\lim_{n\to\infty} f_{n}(x+h) - \lim_{n\to\infty} f_{n}(x) = g(x+h) -g(x)$
$\lim_{n\to\infty} \frac{f_{n}(x+h) - f(x)}{h} = \frac{g(x+h) -g(x)}{h}$
$\lim_{h\to 0} \lim_{n\to\infty} \frac{f_{n}(x+h) - f(x)}{h} = \lim_{h\to 0}\frac{g(x+h) -g(x)}{h}$

A:

There's no reason for $f_n'(x)$ to even converge.  Consider, for instance, $$f_n(x) = \dfrac {\sin(nx)}n$$
Clearly $f_n \to 0$ pointwise on $\mathbb{R}$. However the sequence of derivatives $f^{'}_{n} = \cos(nx)$ does not converge pointwise on $\mathbb{R}$:
$$f^{'}_n(\pi) = (-1)^{n}$$

A:

In order for your assertion to be true, we need one more hypothesis, namely:
$\{f_n'\}$ converges uniformly.
The proof of this can be found on page 152, theorem 7.17 of Principles of Mathematical Analysis by Rudin.
EDIT: For those who don't have the book, the proof is a little involved, so I will just be providing a sketch.
First we notice that we have to prove $\lim_{n\to \infty}\lim_{t\to x}\frac{f_n(t) - f_n(x)}{t-x} = \lim_{t\to x}\frac{f(t) - f(x)}{t-x}$. Wouldn't it be so much easier if we could interchange the limits? Well, as a matter of fact we can, after a bit of justification. Let's look at the theorem that allows us to do so (I won't be proving it over here).
The theorem states that if $f_n$ converges uniformly to a function, then $\lim_{t\to x}\lim_{n\to \infty}f_n(t) = \lim_{n\to\infty}\lim_{t\to x}f_n(t)$
Let $\phi_n(t) = \frac{f_n(t) - f_n(x)}{t-x}$ and $\phi(t) = \frac{f(t) - f(x)}{t-x}$. If we could prove $\lim_{n\to\infty}\phi_n(t) = \phi(t)$ UNIFORMLY, then we could just take the limit as $t\rightarrow x$, interchange the limit and be done.
So how do we prove uniform convergence? Consider $\phi_n(t) - \phi_m(t)$. We need to make this arbitrarily small for large enough n and m. Looking at its form, it would be nice if we could bound the derivative of $f_n - f_m$ and then use the mean value theorem.
Bounding the derivative is easy enough. It's straight from our hypothesis- $\{f'\}$ converges uniformly i.e. $|f_n'(t) - f_m'(t)| < \epsilon$ for large enough m and n. So this fact coupled with MVT should prove uniform convergence of $\phi_n$. Since our hypothesis tells us that $f_n\rightarrow f$ point-wise, we can conclude that $\lim_{n\to \infty}\phi_n(t) = \phi(t)$ uniformly.
Now that we have proved uniform convergence, we can use the theorem that allows us to interchange limits. Taking the limit of $t\rightarrow x$ on both sides and interchanging limits on the left side should give us what we want.
I will leave it to you to fill in the gaps. Anyone, feel free to correct me if I went wrong somewhere.

