Given that $f_n \rightarrow f$ uniformly and that $x_n \rightarrow x$, then I was wondering whether $d (f_n (x_n), f(x)) < \epsilon$, that is, does $f_n(x_n) $converge to$ f(x)$?
Any ideas? Been hitting myself with it. Note that we are NOT given that $f$ is continuous.
On
This is copy-paste from here where you asked for this counterexample.
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 on $[0,1]$ with $x_n\to0$ and $x_n\neq0 , \ \forall n\in\mathbb N$) then $x_n\xrightarrow[n\to\infty]{}0$ but $f_n(x_n)\not\xrightarrow[n\to\infty]{}f(0).$
This is obviously false. Just take $$f_n(x)=f(x)=\begin{cases}1 & x=0 \\ 0 & x \neq 0\end{cases}$$ $$x_n=\frac{1}{n}$$ $$x=0$$