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? 
 A: 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 $\sup_{x\in [0,1]} |f_n(x)|=1$
A: Use the following propisition: "If $(f_n(x))$ converges pointwisley on A to f(x), and there is a sequence $(x_m)_{m\ge 1}$ in A such that $f_n(x_n)-f(x_n)$ does not converges to 0, then the given sequence of functions does not converges uniformly over A".
Now, take $x_n= \frac{1}{n}$.
A: Hint:
Try finding a suitable $N(\epsilon)$ by rearranging as $f_n(x) = \frac{1}{1 + (n-1/x)^2}$.
For non-uniform convergence consider $f_n(1/n)$
A: To prove that $f$ is not uniformly convergent, you need to show that there is an $\epsilon >0$ such that for each $N\in \mathbb N$, there is an $n>N$ and $x_n$ with the property that $\vert f_n(x_n)-f(x_n)\vert >\epsilon$.
So, if we take $\epsilon =1/2$, and $x_n=1/n$, we have $f_n(x_n)=1$ and $f(x_n)=0$ then $\vert f_n(x_n)-f(x_n)\vert =1>\epsilon$ and the claim follows.
An easier way is to note that if $\left \{ f_n \right \}_{n}$ converges uniformly, then $\sup _{x\in \mathbb R}\vert f_n(x)-f(x)\vert\rightarrow 0$, so a routine calculus exercise on max/mins applied to $f_n$ leads again to $x_n=1/n$, from which we get an immediate contradiction.
