Show that $f_n(x_n) \to f(x).$

Question

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 $x$?

Answer 1

Is $∀\epsilon>0$ $∃n_0$ such that $∀n≥n_0:\left\vert f(x_n)−f(x)\right\vert<\epsilon$ the definition of $f(x_n)\to f(x)$ ?

The definition of the limit of numbers $y_n \to y$ is: for all $\epsilon > 0$ there exists $n_0$ such that $n \geq n_0$ implies $\vert y_n - y \vert < \epsilon$. Here, set $y_n = f(x_n)$ and $y = f(x)$. Hence, this is just an application of the definition of limit of numbers.

Furthermore, is $t$ different from $x$?

Well, $t$ ranges over the entire domain (real line, in this case), and in particular, $t$ can be each $x_n$. In the second inequality above, you're bounding $\vert f_n(x_n) - f(x_n) \vert$ (the discrepancy between $f_n$ and $f$ at $x_n$) by the best upper bound of all such discrepancies, i.e., $\displaystyle \sup_{t \in \mathbb{R}} \vert f_n(t) - f(t) \vert$. $t$ has to range over the entire real line, since any one $x_n$ could be anything, in theory.