For $ n \in \mathbb N$
$f_n (x) = \frac{n^2 }{(x^2 +n^2)}$
Let $x \in \mathbb R $. Prove that $f_n (x) \rightarrow 1$ as $n \rightarrow \infty$
My attempt: Let $\epsilon>0 $, then by definition of convergence $\exists N \in \mathbb N$ $\forall n\geqslant N$ : $| \frac{n^2 }{(x^2 +n^2)} -1 |< \epsilon$
$| \frac{n^2 }{(x^2 +n^2)} -1 | = |\frac {x^2}{(x^2 +n^2)}|< \epsilon$
Do not assume $N$ exists, you have to prove it. Your calculations are correct, anyways. Since $x$ is fixed, once you get to the last step you are basically done: you only need to find $N$ so that the last term is smaller than $\epsilon$.
To find $N$, observe that $$ \frac{x^2}{x^2+n^2} < \epsilon \implies \frac{x^2}{\epsilon} < x^2 + n^2 \implies n> \sqrt{\frac{x^2}{\epsilon}-x^2}=\sqrt{\frac{x^2(1-\epsilon)}{\epsilon}}. $$ So $N=\bigg\lceil\sqrt{\frac{x^2(1-\epsilon)}{\epsilon}}\bigg\rceil$ is enough.