Show that
$$\lim \limits_{x \to 1}\sum_{n=1}^{\infty}\frac{n^2x^2}{n^4+x^4}=\sum_{n=1}^{\infty}\frac{n^2}{n^4+1}$$
Justify all steps of your answer by quoting the theorems you are using.
My attempt
Why can't I do the following to get the answer?
$$\lim \limits_{x \to 1}\sum_{n=1}^{\infty}\frac{n^2x^2}{n^4+x^4}=\sum_{n=1}^{\infty}\lim \limits_{x \to 1}\frac{n^2x^2}{n^4+x^4}$$
You can only do that if the series converges uniformly in $x$. Which, luckily for you, it does. On any compact subset of $\mathbb{R}$, say $[-M,M]$, you have that $$\sum_{n=1}^\infty \frac{n^2x^2}{n^4+x^4} \le \sum_{n=1}^\infty \frac{M^2}{n^2}$$.
All you need, in particular, is uniform convergence on a neighborhood of $1$. Can you conclude?
To show $$\lim \limits_{x \to 1}\sum_{n=1}^{\infty}\frac{n^2x^2}{n^4+x^4}=\sum_{n=1}^{\infty}\frac{n^2}{n^4+1}$$
Let $S(x)=\sum_{n=1}^{\infty}\dfrac{n^2x^2}{n^4+x^4}$ and $S=\sum_{n=1}^{\infty}\dfrac{n^2}{n^4+1}$
$\mid \lim \limits_{x \to 1}S(x)-S\mid=\lim \limits_{x \to 1}\mid S(x)-S\mid\leq \lim \limits_{x \to 1}\sum_{n=1}^{\infty}\Big| \dfrac{n^2x^2}{n^4+x^4}-\dfrac{n^2}{n^4+1}\Big|$
$$\Big| \dfrac{n^2x^2}{n^4+x^4}-\dfrac{n^2}{n^4+1}\Big|=n^2\Big| \dfrac{x^2}{n^4+x^4}-\dfrac{1}{n^4+1}\Big|=n^2\Big|\dfrac{n^4(x^2-1)-(x^2+1)(x^2-1)}{(n^4+x^4)(n^4+1)}\Big|\\=n^2|x^2-1|\Big|\dfrac{n^4-x^2-1}{(n^4+x^4)(n^4+1)}\Big|\leq n^2|x^2-1|\dfrac{n^4+3}{n^4\cdot n^4}=|x^2-1|\Big(\dfrac{1}{n^2}+\dfrac{3}{n^6}\Big)$$
Hence $\mid \lim \limits_{x \to 1}S(x)-S\mid\leq \lim \limits_{x \to 1}|x^2-1|\sum_{n=1}^{\infty}\Big(\dfrac{1}{n^2}+\dfrac{3}{n^6}\Big)=\lim \limits_{x \to 1}|x^2-1|\cdot M$
Here $M=\sum_{n=1}^{\infty}\Big(\dfrac{1}{n^2}+\dfrac{3}{n^6}\Big)<\infty$
