Uniform convergence of $f_n(x)=\frac{nx}{(n^2x^2+1)^2}$ Given the function $f_n:[0,1] \to R$
$$f_n(x)=\frac{nx}{(n^2x^2+1)^2}$$
I can show the pointwise convergence $\forall x \in [0,1]: \lim_{n \to \infty} f_n(x) = f(x) = 0$.
To show uniform convergence seems to be much harder to me. I show you what I did below, but it might not be correct (or at least not formal enough). Any help to get a more formal solution would be really appreciate. 
$$\sup_{x \in [0,1]} \left| f_n(x) - f(x) \right| 
= \sup_{x \in [0,1]} \left| f_n(x)\right| 
= \sup_{x \in [0,1]} \left| \frac{nx}{(n^2x^2+1)^2} \right| 
= \sup_{x \in [0,1]} \frac{nx}{(n^2x^2+1)^2} 
= \sup_{x \in [0,1]} \frac{nx}{(nx)^4+2(nx)^2+1}
= \sup_{x \in [0,1]} \frac{1}{(nx)^3+2(nx)+\frac{1}{nx}} 
= \sup_{x \in [0,1]} \frac{1}{(nx)^3+2(nx)+\frac{1}{nx}}
\le \sup_{x \in [0,1]} \frac{1}{(nx)^3+2(nx)}
\le \frac{1}{(n)^3+2(n)}
\underrightarrow{n \to \infty} 0$$
 A: The convergence $f_n\to0$ is not uniform since $f_n(1/n)=1/4$ for every $n$.
A: To find 
$$ \sup_{x \in [0,1]} \frac{nx}{(n^2x^2+1)^2}, $$
you need to find the maximum of the above function
$$ \frac{nx}{(n^2x^2+1)^2} $$ 
for $x\in[0,1]$. You can use applications of the derivative to find the max and min of a function. 
A: Before deciding if a given sequence is uniformly convergent or not, it's usually a good idea to look at the graphs of the $f_n$.  
Below are the graphs of $\color{darkblue}{f_1}$, 
$\color{maroon}{f_2}$, 
$\color{darkgreen}{f_5}$, 
$\color{pink}{f_8}$, 
$\color{cyan}{f_{10}}$, and 
$\color{yellow}{f_{40}}$ (sorry if the colors are too vibrant).

From this you might surmise that the convergence isn't uniform, as each $f_n$ seems to "peak" with value $1/4$, and since the pointwise limit is the zero function. 
And indeed the convergence of $(f_n)$ to the zero function isn't uniform: you could show that each $f_n$ assumes the value $1/4$ (and even that this is the maximum value) using the methods of the other answers.
A: You can find maximum of $ \; \dfrac{u}{u^4+2u^2+1}$
