= Solution
A sequence of functions $f_n:A\to\mathbb R$ <uniform convergence>[converges uniformly] to $f$ when, for every $\varepsilon>0$, there is an $N$ such that
$$
n\geq N\quad\Longrightarrow\quad |f_n(x)-f(x)|<\varepsilon
$$
for every $x\in A$. Equivalently, $\lVert f_n-f\rVert_\infty\to0$ in the <supremum norm>.
\b[Yes.] Fix $x_0\in A$ and $\varepsilon>0$. Uniform convergence gives $N$ such that $|f_N(x)-f(x)|<\varepsilon/3$ for every $x\in A$. Since the <continuous function> $f_N$ is continuous at $x_0$, there is a $\delta>0$ such that $|x-x_0|<\delta$ implies $|f_N(x)-f_N(x_0)|<\varepsilon/3$. The <triangle inequality> then gives
$$
|f(x)-f(x_0)|
\leq |f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|
<\varepsilon.
$$
This proves the <uniform limit theorem>: $f$ is continuous.
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