$f_n(x)=\frac{x^2}{x^2+(1-nx)^2}$ is uniformly convergent on $[\delta,1]$ for all $\delta \in (0,1)$ I would like to show that the sequence of functions $\{f_n\}$ defined by $f_n(x)=\frac{x^2}{x^2+(1-nx)^2}$ is uniformly convergent on $[\delta,1]$ for all $\delta \in (0,1)$. 
I will fix a $\delta \in (0,1)$. I see that $f_n(x)\rightarrow f(x)=0$ for each $x\in [\delta,1]$ (pointwise convergence).
Given an $\epsilon >0$ I need to find an $N \in \mathbb{N}$ such that for all $n \geq N$
$$|f_n(x)-f(x)|=\mid\frac{x^2}{x^2+(1-nx)^2}\mid <\epsilon$$
for all $x\in [\delta,1]$. I am having trouble finding this $N$. I know I can bound $f_n$ above by $\frac{1}{n^2-2n+2}$, but that is not really helping me. I've even tried looking at $f_n'$, but that was treacherous. I have hit a wall. I would really appreciate some help. Thank you :-)
 A: The key is to compute $f_n'(x)$. You'll find that on $[0,1]$ the derivative is initially positive, becomes zero at $x=\frac1n$, then is negative after that. So $|f_n-f|$ (which equals $f_n$, since $f_n$ is nonnegative) is maximized at $x=\frac1n$ and decreases from that point.
Since $\frac1n\to 0$ as $n\to\infty$, this means that for all large enough $n$, we have $f_n$ is positive and decreasing throughout the interval $[\delta,1]$, hence for all large $n$ the max value of $f_n$ is achieved at the left endpoint $x=\delta$:
$$
\sup_{x\in[\delta,1]}\left|f_n(x)-f(x)\right| = f_n(\delta).
$$
Your job is to show that $f_n(\delta)\to0$ as $n\to\infty$.
A: First we can restrict to working with those $n's$ that are greater than $\dfrac{1}{\delta}$. This means $\dfrac{1}{n} < \delta$. Thus $\dfrac{n}{1+n^2} < \dfrac{1}{n} < \delta$. Then the denominator of $f_n(x)$ which is the function $g_n(x)= x^2+(1-nx)^2 = (1+n^2)x^2-2nx+1$ has derivative $g_n'(x)= 2(1+n^2)x-2n > 0$ on $[\delta, 1]$ by using the observation above. Thus $g_n(x) \ge g_n(\delta) = \delta^2+(1-n\delta)^2$. Thus since $0 < x \le 1$, $f_n(x) \le \dfrac{1}{\delta^2+(1-n\delta)^2}< \epsilon\iff n\delta - 1 > \sqrt{\dfrac{1}{\epsilon} - \delta^2} \implies n > \dfrac{1}{\delta}\cdot \left(1+\sqrt{\dfrac{1}{\epsilon}-\delta^2}\right)= N$.
