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Let $(f_n)$ be a sequence of continuous functions on $[a,b]$ that converges uniformly to $f$ on $[a,b]$. Show that if $(x_n) \subseteq[a,b]$ and if $x_n \rightarrow x$, then $\lim f_n(x_n) = f(x) $

My solution: Let $\epsilon > 0$. Take $N \in \mathbb{N}$ such that

$|f_n(x) - f(x)| < \frac{\epsilon}{2}, \forall n>N \text{ and } \forall x \in [a,b].$

Now, since $(f_n)$ is continuous, take $\delta > 0$ such that:

$|x_n - x| < \delta \Rightarrow |f_{N+1}(x_n) - f_{N+1}(x)| < \frac{\epsilon}{2}.$

Now, if $|x_n - x| < \delta$, we must have that:

$|f_n(x_n) - f(x)| \leq |f_n(x_n) - f_n(x)| + |f_n(x) - f(x)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

Is this correct?

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In the last sentence how do you know that $|x_n-x|<\delta \Rightarrow |f_n(x_n)-f_n(x)|<\frac{\epsilon}{2}?$ Also where did you use uniformly convergence? Note that $x$ is fixed. Your proof is not correct.

For a correct proof use that $f_n \xrightarrow[n\to\infty]{}f$ uniformly, $f$ is continuous and split $|f_n(x_n)-f(x)|\leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|$ for all $n\geq N$ where $N$ is such that $|f_n(y)-f(y)|<\dfrac{\epsilon}{2} \text{ and } |f(x_n)-f(x)|<\dfrac{\epsilon}{2}$ for all $n\geq N$ and $y \in [a,b].$

The assumption of $(f_n)$ being continuous is necessary as the following (counter) example shows.

Consider the sequence of functions $(f_n(x))_{n\in\mathbb N}$ on $[0,1]$ defined by $f_n(x)=\begin{cases}1 \ \ , \ \ \ x=0\\ \\ \dfrac1n \ \ , \ \ \ x\in \left(0,1\right]{}\end{cases}.$ The sequence $f_n$ converges uniformly to the function $f(x)=\begin{cases}1 \ \ , \ \ \ x=0\\ \\ 0 \ \ , \ \ \ x\in \left(0,1\right]{}\end{cases}$ and if $x_n=\dfrac1n$ (or any sequence with $x_n\to0$ and $x_n\neq0 , \ \forall n\in\mathbb N$) then $x_n\xrightarrow[n\to\infty]{}0$ but $f_n(x_n)\not\xrightarrow[n\to\infty]{}f(0).$

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    @user43901: Oops, wrong counter-example. I'll update my answer.2013-03-15