Definition: A sequence of function $f_1, f_2, \ldots$, converges uniformly on a set $E$ to a function $f$ if for every $\eps > 0$ there exists $N$ such that $n > N$ implies
\[\abs{f_n(x) - f(x)} < \eps\]for all $x \in E$.
Corollary: $\,f_n \to f$ uniformly on $E$ if and only if
\[\sup_{x \in E} \abs{f_n(x) - f(x)} \to 0.\]Theorem: Suppose $f_n \to f$ uniformly, with $f_n$ continuous for all $n$. Then $f_n(\x) \to f(\x_0)$ as $\x \to \x_0$.
Proof: Let $\eps > 0$. For any sequence $\x_m \to \x_0,$
\[\begin{alignat*}{2} \tag*{$\tcirc{1}$} \exists N: n > N \implies \sup_x \abs{f_n(\x) - f(\x)} &< \tfrac{\eps}{2} &\hspace{4em}& \text{Def. Uniform convergence} \\ \tag*{$\tcirc{2}$}\exists M: m > M \implies \abs{f(\x_m) - f(\x_0)} &< \tfrac{\eps}{2} && f \text{ is } \href{continuous.html}{\text{continuous}}\text{; see corollary below} \end{alignat*}\]Thus, for all $n > N$ and $m > M$, we have
\(\begin{alignat*}{2} \abs{f_n(\x_m) - f(\x_0)} &= \abs{f_n(\x_m) - f(\x_m) + f(\x_m) - f(\x_0)} &\hspace{4em}& \\ &\leq \abs{f_n(\x_m) - f(\x_m)} + \abs{f(\x_m) - f(\x_0)} && \href{norm.html}{\text{Triangle inequality}} \\ &< \sup_x \abs{f_n(\x) - f(\x)} + \tfrac{\eps}{2} && \tcirc{2} \\ &< \eps && \tcirc{1} \end{alignat*}\)
Additional results
Theorem: Suppose $f_n \to f$ uniformly on $E$ and that $\lim_{x \to x_0} f_n(x)$ exists for all $n$. Then for any limit point $x_0$ of $E$,
\[\lim_{x \to x_0} \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \lim_{x \to x_0} f_n(x).\]Proof: Rudin 7.11.
Corollary: If ${f_n}$ is a sequence of continuous functions on $E$ and if $f_n \to f$ uniformly on $E$, then $f$ is continuous on $E$.