Disproving uniform convergence Given a sequence of functions $f_n$ which is known to be pointwise convergent how would you go about showing that is not uniform convergent?   
The particular example I'm working with is $f_n (x)=\frac {nx} {x^2+n^2}$ and I've tried using Theorem 7.9 from Rudin (uniform convergence $\iff M_n=(\sup |f_n(x)-f(x)|) \rightarrow 0$ as $n\rightarrow \infty$) but using that I found $M_n$ to be $\frac {\sqrt n} {1+n}$ by maximising $f_n$ but this gives the wrong result.
edit:
$f:\mathbb R \rightarrow \mathbb R $, sorry about that
 A: It looks like you've mis-computed $M_n$.
In general, let $g$ be a function continuous at $0$, and define $g_n(x)= g(x/n)$.
Then $g_n(x) \rightarrow g(0)$ as $n\rightarrow\infty$ for all $x$.
But $M_n=\sup_x |g_n(x)-g(0)| = \sup_x |g(x)-g(0)|=M_1$ for all $n$, so $M_n\rightarrow 0$ if and only if $M_1=0$ - that is, $g$ is constant.  So $g_n$ does not converge uniformly to $g(0)$ unless $g$ is constant.
In your case, $f_n(x) = \frac {x/n}{1+(x/n)^2}$
A: There exist a criterion, in some books is an exercise:
Let $(f_n)$ a sequence of real valued functions on $D$. If there exist $(x_n)$ and $(y_n)$ sequences in $D$, such that
$$|x_n-y_n|\to 0$$
but
$$|f_n(x_n)-f_n(y_n)|\not \to 0$$
then, the sequence $(f_n)$ do not converges uniformly on $D$.
In your particular case, take
$$x_n=n\quad \text{ and }\quad y_n=n+\frac{1}{n}.$$
A: The uniform limit of continuous functions is again continuous. 
So if $\{f_n\}$ are continuous and $f_n\to f$ pointwise on $I$, but $f$ is not continuous on $I$, then $f_n$ cannot converge uniformly to $f$.
Of course, this is only helpful in the case that $f$ is not continuous...
