$f_n(x_n) \rightarrow f(x) $ by uniform convergence

Solution 1:

I will give a proof using a slightly different version of the triangle inequality than the one used by OP namely

$$|f_n(x_n) - f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)|$$

We start off by fixing $\epsilon > 0$. From the definition of uniform convergence there exist a $N$ s.t. if $n>N$ the first term is less than $\epsilon / 2$.

The uniform convergence of $f_n\to f$ and the fact that $f_n$ is continuous it follows that $f$ is continuous. This implies that there exist a $\delta>0$ s.t. if $|x-x_n|<\delta$ then $|f(x_n) - f(x)| < \epsilon/2$. Finally since $x_n\to x$ there exist a $M$ s.t. if $n>M$ we have $|x-x_n| < \delta$.

Putting it all togeather we find that if $n>\text{max}(N,M)$ then

$$|f_n(x_n) - f(x)| < \epsilon/2 + \epsilon/2 = \epsilon$$