Let $f_n$ continuous,so that $f_n \to f$ uniformly.Let $x_n$ be a sequence of real numbers,such that $x_n \to x$.Show that $f_n(x_n) \to f(x).$
$f_n$ continuous and $f_n \to f$ uniformly,so $f$ is continuous and $\sup_{t \in \mathbb{R}} \{ |f_n(t)-f(t)|\} \to 0$
$f$ is continuous and $x_n \to x$,so $f(x_n) \to f(x)$ ,so $\forall \epsilon>0$ $\exists n_0 $ such that $\forall n \geq n_0: |f(x_n)-f(x)|< \epsilon$
$|f_n(x_n)-f(x)|=|f_n(x_n)-f(x_n)+f(x_n)-f(x)| \leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)| \leq \sup_{t \in \mathbb{R}} |f_n(t)-f(t)|+|f(x_n)-f(x)| \to 0$
Is $\forall \epsilon>0$ $\exists n_0 $ such that $\forall n \geq n_0: |f(x_n)-f(x)|< \epsilon$ the definition of $f(x_n) \to f(x)$ ?
Furthermore, is $t$ different from $x$?