Let $(f_n)$ be a sequence of continuous functions such that $f_n\to f$ uniformly on $X.$ Show that $ f_n(x_n) \to f(c) \; $ for all sequences $ \;x_n\to c.$
I know that $f_n \to f$ uniformly where $f_n$ are continuous $\rightarrow f$ continuous
I also know that $$\lim_{n\to \infty}\lim_{x\to c} f_n(x)= \lim_{x\to c} \lim_{n\to \infty} f_n(x).$$
So can I prove it like this:
$$\lim_{n\to \infty} f_n(x_n) = \lim_{n\to \infty} \lim_{x\to c} f_n(x) =\lim_{x\to c} \lim_{n\to \infty} f_n(x) = \lim_{x\to c} f(x) = f(c)?$$
A standard $\varepsilon/3$-style argument works here (only we need $\varepsilon/2$ in this case). Let $\varepsilon>0.$ Because $f_n\to f$ uniformly on $X,$ there exists $N>0$ such that for every $n\ge N$ and $x\in X$ we have $$|f_n(x)-f(x)|<\frac{\varepsilon}{2}. $$ Similarly, because $f$ is continuous (the limit of continuous functions is continuous if the convergence is uniform), there exists $\tilde{N}>0$ such that for every $n\ge \tilde{N},$ we have $|f(x_n)-f(c)|<\dfrac{\varepsilon}{2}.$ Let $N'=\max(N,\tilde{N}),$ and assume $n\ge N'.$ Then \begin{align*} |f_n(x_n)-f(c)|&= |f_n(x_n)-f(x_n)+f(x_n)-f(c)|\\ &\le |f_n(x_n)-f(x_n)|+|f(x_n)-f(c)|\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\ &=\varepsilon, \end{align*} and we are done.