How to show that $f_{n}(x)=\frac{x^{2}}{x^{2}+(1-nx)^{2}},\,0\leq x\leq1 $ is not uniformly convergent The problem says:

$f_{n}(x)=\frac{x^{2}}{x^{2}+(1-nx)^{2}},\,0\leq x\leq1$ is convergent pointwise to $0$ but not uniformly.

So, I show that $f_n \rightarrow f$(pointwise) on $[0,1]$.
And I tried:
If I take $\epsilon=1/2N$, then
$$\frac{1}{2N}\leq\frac{1}{N}\left|\frac{1}{2-Nx}\right|=\left|\frac{x^{2}}{2Nx-N^{2}x^{2}}\right|=\left|\frac{x^{2}}{1-(1-Nx)^{2}}\right|\leq\left|\frac{x^{2}}{x^{2}-(1-Nx)^{2}}\right|$$
Thus, it is not convergent to $0$ uniformly on $[0,1]$.
But it seems to be wrong since $\epsilon$ is dependent on N. 
How can I fix it? 

Note that $f_n(x)$ attains its maximum value when $nx=1\implies x=\dfrac{1}{n}$.
Also for $f_n\to f$ uniformly $\sup_{x\in [0,1]} |f_n(x)|\to 0$ which is false here as $\sup_{x\in [0,1]} |f_n(x)|=1$
