TITLE: Pointwise convergence of sequence $(f_n)_n$ of functions to $f$ and changing limits
QUESTION [7 upvotes]: My analysis notes contains the following question: if $(f_n)_n$ is a sequence of functions of $A \subset \mathbb{R} \to \mathbb{R}$ and $a \in \mathbb{R} \cup \{-\infty, +\infty\}$ an accumulation point in $A$. Assume that for all $n$, $\lim_{ x \to a}f_n(x) = L_n$ exists and is finite. Suppose $(f_n)_n$ converges pointwise to $f:A \to \mathbb{R}$. 
1) Does $\lim_{x \to a} f(x)$ exists?
2) Does $\lim_{n \to \infty} L_n$ exists?
3) Can we change limits (in case both limits exist)?

If know all questions should be no. I have found examples for 1) and 3). However, I can not find an example for 2). I know I should look for a sequence of functions which does not converge uniformly. Any hints would be appreciated.

My solutions to 1) 
Define $$f_n: \mathbb{R} \to \mathbb{R}: x \mapsto \begin{cases} -1 &\text{ if } x \leq -1/n\\ nx & \text{ if } -1/n < x < 1/n\\ 1 &\text{ if } x \geq 1/n\end{cases}.$$ This functions converges to the function $f$ which equals to -1 for $x > 0$, 0 for $x = 0$ and $1$ for $x > 0$. The limit in zero does not exist.
and 3):
define $$f_n: \mathbb{R} \to \mathbb{R}: x \mapsto \frac{(nx)^2}{1 + (nx)^2}.$$ This sequence converges to $f$ which equals 1 everywhere, except for $x = 0$, where it equals to $0$. We have that 
$$1 = \lim_{x \to 0} \lim_{n \to \infty} f_n(x) \neq \lim_{n \to \infty} \lim_{x \to 0} f_n(x) = 0$$

REPLY [2 votes]: Let $\{q_1,q_2,\dots\}$ be an ordering of $\mathbb{Q}$. Define $f : \mathbb{Q} \to \mathbb{R}$ as follows:
$$f_n(x)=\begin{cases}
0 &: x\in \{q_i : 1 \leq i \leq n\}\\
n &: \text{otherwise}\end{cases}$$
Then, $\{f_n\}$ converges to $0$ pointwise and for $a=0$, $L_n=n \, \forall n \in \mathbb{N}$. Clearly, it contradicts $2$.