Conditions which assure $f_n(a_n) \to f(a)$ assuming $f_n \to f$ and $a_n \to a$ For a larger proof I'd need that if $\lim_{n\to\infty} a_n = a$ then $$
  \lim_{n\to\infty} \left(1+\frac{a_n}{n}\right)^n = e^a \text{.}
$$
Which made me wonder what conditions are sufficient for $$
  \lim_{n\to\infty} f_n(a_n) = f(a)
$$
to hold, assuming that $\lim_{n\to\infty} f_n = f$, $\lim_{n\to\infty} a_n=a$, and obviously that $f$ is continuous at $a$.
If one assumes uniform convergence of $f_n \to f$ then things should be simple, because one can then find an $N$ and a $\delta$ for every $\epsilon$ such that $f_n([a-\delta,a+\delta]) \in [f(a)-\epsilon,f(a)+\epsilon]$ for all $n \geq N$.
Edit: What follows turned out to be wrong, see answers
It seems, however, that this actually only requires uniform convergence on some (wlog closed, hence compact!) interval around $a$. Which doesn't appear to be a restriction at all, since pointwise convergence on a compact set implies uniform convergence if I'm not mistaken (if the limit function is continuous, that is).
Which would mean that no further restrictions are necessary, and $f$ being continous is both necessary and sufficient. Does anyone see a hole in this (quite sketchy) proof?
 A: Here is another example which shows that pointwise convergence of a sequence of functions $f_1$, $f_2$, $f_3$, $\dots$ to $f$ on a compact set $K$ does not imply uniform convergence, even if the limit function $f$ is continuous and all functions in the sequence are continuous: take $K=[0,1]$ and
$$
f_n(x)=\left\{\begin{array}{ll} 0, & x=0,\\
nx, & 0<x\le \frac 1n,\\
2-nx, & \frac 1n < x\le \frac 2n,\\
0, & \frac 2n < x\le 1.
\end{array}\right.
$$
Then, for all $x\in [0,1]$, $f_n(x)\to 0=:f(x)$ as $n\to\infty$, but the convergence is not uniform.  Taking $a_n=\frac 1 n$ and $a=0$, this also shows that pointwise convergence of $f_n$ to $f$ and convergence of $a_n$ to $a$ does not necessarily imply that $f_n(a_n)$ converges to $f(a)$, since here $f_n(a_n)$ is always $1$ but $f(a)=0$.
If $f$ and $f_1$, $f_2$, $\dots$ are all continuous and real-valued and either the sequence $f_1(x)$, $f_2(x)$, $f_3(x)$, $\dots$ is monotonically increasing for all $x$ or it is monotonically decreasing for all $x$, you can use Dini's theorem to conclude that the convergence is uniform on $K$.
