<s>[INST] A problem with proving using definition that $\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1$ 
Prove using the definition that: $$\displaystyle\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1 $$

What I did:
Let $\epsilon >0$, finding $N$: $\mid\frac {n^2-1}{n^2+1}-1\mid=\mid\frac {-2}{n^2+1}\mid\le\mid\frac {-2}{n^2}\mid=\frac {2}{n^2}\le\frac {2}{n}$ So $N=\frac 2 {\epsilon}$.
So for all $n>N$ we'll want to show that: $|a_n-l|<\epsilon \Rightarrow \mid\frac {n^2-1}{n^2+1}-1\mid=\mid\frac {-2}{n^2+1}\mid<\mid\frac {-2}{N^2} \mid=\mid\frac {-2}{\frac 1 {\epsilon^2}} \mid=\mid -2\epsilon^2 \mid=2\epsilon^2$
But that isn't smaller than $\epsilon$ so I'm doing something wrong here and I don't know what...
 [/INST] $$\left|\frac{n^2-1}{n^2+1}-1\right|=2\frac1{n^2+1}<\epsilon\iff n^2+1>\frac2\epsilon\iff n>\sqrt{\frac2\epsilon-1}$$
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