Let $X$ be a subset of $\mathbb R$, $(f_n)_n$ be a sequence of functions on $X$. Assume that the sequence $f_n$ converges pointwise towards a function $F$ and uniformly on every compact subsets of $X$. Let $a$ be a point of $\overline X$, the adherence of $X$.
Assume that for each $n\in\mathbb N$, the function $f_n$ admits a limit $l_n$ at the point $a$. Can one assert that the function $F$ admits a limit at $a$?
If (1) is false, if, in addition, one assumes that the sequence $(l_n)_n$ converges towards an element $u\in\mathbb R\cup\{\pm\infty\}$, can one conclude that $F$ admits $u$ as limit at $a$ then?
Thanks in advance for any answer.
Consider the example: $X=[0, 1), f_n(x) = x^n, F(x)=0$. The convergence is uniform on any $[0, b], b<1$ since $|f_n(x)-F(x)|\leq b^n$.
Let $a=1\in \overline X$. Then $\lim_{x\to 1} f_n(x)=1$, but $\lim_{x\to 1}F(x)=0$.
Here is an even cooler example contradicting 1). Let $X=(0, 1]$. Consider \begin{align} f_n(x)&=\frac{x}{x^2+\frac{1}{n}},\\ F(x) &=\frac 1 x. \end{align} Then $f_n\to F$ uniformly on any $[\delta, 1], \delta>0$, since $$ |f_n(x)-F(x)| = \frac{\frac{1}{n}}{(x^2+\frac{1}{n})x} \leq \frac {1} {\delta^3 n}. $$ Now let $a=0\in \overline{X}$. Then $$ \lim_{x\to a}f_n(x)=0,\quad\forall n, $$ but $\lim_{x\to a}F(x)=\infty$, or does not exist.