Htykuut

I recently read this theorem in real analysis:(Actually a corollary to a theorem)

{$f_m$} is a sequence of continuous functions defined on $D$ such that $f_m$$to$$f$ uniformly on $D$ then for every sequence {$x_m$} in $D$ satisfying $x_m$$to$$x$ then $f_m(x)$$to$$f(x)$.

please correct me if I wrote this wrong. The doubt is that I am unable to use it in a problem:
$$f_m(x)=dfrac{mx}{1+mx}qquad xin[0,1].$$
I want to prove that this is not uniformly convergent, but I am not able to understand what is $x_m$. How does it depend on $m$? Does it even depend on $m$?
If yes then how to prove that $f_m(x)$ does not converge to $f(x)$ when $x_m$ tends to $x$(to show that it is not uniformly convergent).(also, is there $x_m$ or $x$ inside $f_m(cdot)$ . Basically, how to approach this problem, please explain stepwise because I need to get the feeling of this theorem.

Another way to show $f$ is not uniformly convergent.
$$f_m=begin{cases}0 &x=0\dfrac{mx}{1+mx} &xneq0end{cases}$$ $:$For $xneq0:$ $f_m(x)=dfrac{mx}{1+mx}=dfrac{x}{frac1m+x}$. $:$ Now if $mtoinftyimplies f_m(x)to1$.

So $$f_mtobegin{cases}0 &x=0\1 &xneq0end{cases}=f$$

and $f$ is clearly not continuous. Hence the convergence is not uniform.

$rule{17cm}{1pt}$

$0neq x_m=dfrac1mto0implies:forall:mquad f_m(x_m)=dfrac12notto f(0)=0$. So the convergence is not uniform.