$\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:
$\lim_{n\to\infty} f_{n}(x) = g(x)$
$-\lim_{n\to\infty} f_{n}(x) = -g(x)$
$\lim_{n\to\infty} f_{n}(x+h) - \lim_{n\to\infty} f_{n}(x) = g(x+h) -g(x)$
$\lim_{n\to\infty} \frac{f_{n}(x+h) - f(x)}{h} = \frac{g(x+h) -g(x)}{h}$
$\lim_{h\to 0} \lim_{n\to\infty} \frac{f_{n}(x+h) - f(x)}{h} = \lim_{h\to 0}\frac{g(x+h) -g(x)}{h}$

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 $\mathbb{R}$:
$$f^{'}_n(\pi) = (-1)^{n}$$
