Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of complex-valued functions on the subset $E$ of a metric space $(X,d)$. We all know that if $p$ is a limit point of $E$ and $f_n$ converges uniformly to a limit function $f$ on $X$, then limits are interchangeable, namely:$$\lim \limits _{x\to p}\lim \limits _{n\to \infty}f_n(x)=\lim \limits _{n\to \infty}\lim \limits _{x\to p}f_n(x),$$provided that $\lim \limits _{x\to p}f_n(x)$ exists eventually (that is, for every $n$ greater than a certain $N\in \mathbb{N}$).
Question
Does the same result of interchangeability hold also when we weaken the statement by making $f_n$ converge locally uniformly to $f$ on $E$?
My attempt
It is clear that if $p$ is a limit point of $E$ which belongs to $E$ then the statement is true for $f_n$ converges uniformly to $f$ on a neighbourhood of $p$. However, this made me think abou the case in which $p$ does not belong to $E$. Indeed, in that case, it seems to me that the statement does not hold becuase we can get arbitrarily close to $p$ with the neighbourhoods in which $f_n$ converges uniformly, however we can never be sure that $p$ is actually in such neighbourhoods and, hence, we can never be sure that local uniform continuity can be used to prove the statement.
I would like to know if this seems right or not and if there is a counter-example that could show that the statement does not always hold.
As always, any comment or answer is welcome and let me know if I can explain myself clearer!
An easy counterexample is the following one: $$E=(0,1), \mbox{ } f_n(x)=x^n$$ If $K$ is an arbitrary compact into $(0,1)$ we have $$\sup_K x^n=(\sup K)^n \rightarrow 0$$ and therefore $f_n$ converges locally uniformly to $0$ in $E$. However: $$\lim_{x \rightarrow 1^-} \lim_{n \rightarrow \infty} x^n = 0 \neq 1 = \lim_{n \rightarrow \infty} \lim_{x \rightarrow 1^-} x^n$$