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\noindent \textbf{Theorem:} Suppose $f_n \to f$ uniformly, with $f_n$ continuous for all $n$.  Then $f_n(\x) \to f(\x_0)$ as $\x \to \x_0$.

\begin{proof}
  Let $\eps > 0$.

  %% Note: &\hspace{XX}& Controls the spacing between columns. The 

\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 \delta: \norm{\x - \x_0} < \delta \implies \abs{f(\x) - f(\x_0)} &< \tfrac{\eps}{2} && f \text{ is continuous}
\end{alignat*}

Therefore, for $n > N$ and any $\x \in N_\delta(\x_0)$, we have
\begin{alignat*}{2}
\abs{f_n(\x) - f(\x_0)} &= \abs{f_n(\x) - f(\x) + f(\x) - f(\x_0)} \\
 &\leq \abs{f_n(\x) - f(\x)} + \abs{f(\x) - f(\x_0)} &\hspace{4em}& \text{Triangle inequality} \\
 &< \sup_x \abs{f_n(\x) - f(\x)} + \tfrac{\eps}{2} && \tcirc{2} \\
 &< \eps && \tcirc{1}
\end{alignat*}
\end{proof}

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