Proving pointwise convergence

Question

I'm having trouble proving: $$f_n(x) = \frac{1-n^2x^2}{(1+n^2x^2)^2}$$

converges pointwise to zero for $x \in \,[-1,0)\cup(0,1]$.

My attempt for finding : $N(\epsilon,x)$ \begin{align*} \left|\frac{1-n^2x^2}{(1+n^{2}x^{2})} - 0\right|<\epsilon &\iff \frac{|1-n^2x^2|}{(1+n^2x^2)^2} < \epsilon\\ \end{align*}

Since, \begin{align*} \frac{|1-n^2x^2|}{(1+n^2x^2)^2} &< \frac{|1-n^2x^2|}{n^4x^4}\\ \end{align*}

we can then find $\epsilon$ such that: \begin{align*} \frac{|1-n^2x^2|}{n^4x^4}<\epsilon \end{align*}

I get stuck after this step. Does anyone know how to prove this without using $$\lim_{n \rightarrow \infty}f_n(x)$$

Answer 1

HINT

Let $x \in [-1,0) \cup (0,1]$ and $\epsilon>0$

\begin{align*} |f_n(x)| \leq \frac{|1-n^2x^2|}{n^4x^4}\leq\frac{1}{n^4x^4}+\frac{n^2x^2}{n^4x^4}=\frac{1}{n^4x^4}+\frac{1}{n^2x^2} \end{align*}

Exists $n_1 \in \mathbb{N}$ such that for all n$\geq n_1$ we have $\frac{1}{n^4x^4} \leq \epsilon\Rightarrow n_1 \geq \frac{1}{\sqrt[4]{\epsilon}x^4}$

Exists $n_2 \in \mathbb{N}$ such that for all n$\geq n_2$ we have $\frac{1}{n^2x^2} \leq \epsilon\Rightarrow n_2 \geq \frac{1}{\sqrt{\epsilon}x^2}$

Take $N \geq \max\{n_1,n_2\}$ and you are done.