let $f_n: R \rightarrow R$ be a sequence of continuous functions which converge uniformly to $f: R \rightarrow R$. Let $(x_n)$ be a sequence of real numbers which converges to $x \in R$. Show that $f_n(x_n) \rightarrow f(x)$.
So far this is my attempt at a solution but I wonder whether I can provide a shorter or more insightful proof:
1) $|f_n(x_n)-f(x)| \leq |f_n(x_n)-f_n(x)| + |f_n(x)-f(x)|$ and I argue that we can show that the expression on the right approaches zero as $n \rightarrow \infty$.
It's given that $\lim_{n \to +\infty}|f_n(x)-f(x)| = 0$ since $f_n \rightarrow f$ uniformly.
2) $\forall n, f_n(x)$ is continuous so $\forall \epsilon > 0 \exists \delta > 0, |f_n(x_n)-f_n(x)|<\epsilon \text{ when } |x_n-x|<\delta$.
Clearly, $\lim_{n \to +\infty}|f_n(x)-f_n(x)| = 0$ since $\lim_{n \to +\infty}|x_n-x| = 0$.
Q.E.D.