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...
 A: You were almost there, but near the end you strayed from your correct scratch work.

$$|a_n-l|<\varepsilon \Rightarrow \left|\frac {n^2-1}{n^2+1}-1\right|=\left|\frac {-2}{n^2+1}\right|\color{red}<\left|\frac {-2}{N^2} \right|=\left|\frac {-2}{\frac 1 {\varepsilon^2}} \right|=\left| -2\varepsilon^2 \right|=2\varepsilon^2.$$

The red inequality is not incorrect, but by using it you're discarding the work you've done before.
Along the line of what you did before, you should have done:
$$\left|\frac {-2}{n^2+1}\right|<\dfrac 2n<\dfrac 2 N=\varepsilon.$$
Apparently you allow for real $N$ in your definition. If you wanted a natural number $N$, it would suffice to take a natural number larger than $\dfrac 2\varepsilon$, (they exist).
A: $$\left|\frac{n^2-1}{n^2+1}-1\right|=2\frac1{n^2+1}<\epsilon\iff n^2+1>\frac2\epsilon\iff n>\sqrt{\frac2\epsilon-1}$$
A: Hint:-
$\dfrac{n^2-1}{n^2+1}=1-\dfrac{2}{n^2+1}<1-\dfrac{1}{n^2}$
A: Hint : Dive denominator and numerator by the highest appearing power, here $n^2$
