Limit of a locally uniformly convergent sequence of continuous functions

I have two questions:
1. I know that the uniform limit of a continuous functions is continuous. But I'm wondering whether this is true if the convergence is locally uniform. That is the uniform convergence is true within any bounded interval.
2. Also I'm confused with the uniqueness of the limit.
For example:
Define
$$f_n:[0,1]\rightarrow[0,1]$$ such that
$$f_n(x) = \begin{cases} 1-nx &, 0 \leq x\leq \frac{1}{n} \\ \\0 &, \frac{1}{n} \leq x\leq 1 \end{cases}$$ Define $$f:[0,1]\rightarrow [0,1]$$ to be zero function.
Define $$h:[0,1]\rightarrow [0,1]$$ to be $$h(x)=\begin{cases} 1 &, x=0\\ \\ 0 &, x\neq 0 \end{cases}$$
and we can prove that: $$\forall x,y\in (0,1]$$, $$f_n$$ converges uniformly to $$f$$ as well as to $$h$$ within $$[x,y]$$ (That is locally uniformly). Please point out the mistake I have done

Since continuity is a local property (that is, in order to determine whether $$f$$ is continuous at $$a$$, all that matters is how $$f$$ behaves near $$a$$), yes, local uniform convergence preserves continuity.
Concerning your other question, $$(f_n)_{n\in\mathbb N}$$ neither uniformly to no function and converges pointwise to $$h$$.
• The limit is unique (if it exists) because, for each $x$ in the domain of the $f_n$'s, $f(x)=\lim_{n\to\infty}f_n(x)$ and the limit of a sequence of numbers is unique (again, if it exists). – José Carlos Santos Dec 8 '18 at 15:06