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
$$1 = \lim_{x \to 0} \lim_{n \to \infty} f_n(x) \neq \lim_{n \to \infty} \lim_{x \to 0} f_n(x) = 0$$
Best Answer
Let $(f_n)_{n \geq 1} : \mathbb{R} \to \mathbb{R}$ be defined as $f_{2n} = \chi_{[2n,+\infty)}$ and $f_{2n-1} = \chi_{(-\infty, 2n-1]}$. Now, if $x \in \mathbb{R}$, there exists $k \in \mathbb{N}$ such that $x \in [-k,k]$ and so if $n \geq k$, we have that $f_n(x) = 0$. Therefore we have pointwise convergence.
Moreover, if $a = \infty$ and $n \in \mathbb{N}$, then $\lim_{x \to \infty}f_n(x)$ always exists,
$$ \lim_{x \to \infty}f_n(x) = \cases{0 \quad n \text{ is odd} \\ 1 \quad n \text{ is even}} $$
However, it is clear from here that $(L_n)_{n\geq 1}$ does not converge, singe taking odd and even terms we have two subsequences converging to different values.
As a side note, a sufficient condition for $(2)$ to hold is that $a \in \mathbb{R}$ and $f_n$ continuous at $a$ for all $n$, in which case $L_n \to f(a)$. If so, we would have that
$$ |f(a) - L_n| \leq |f(a) - f_n(a)| + |f_n(a)-f_n(x)| + |f_n(x) - L_n| $$
Taking limit of $x \to a$, we have that
$$ |f(a) - L_n| \leq |f(a) - f_n(a)| \xrightarrow{n \to \infty} 0 $$
as claimed.