Prove the sequence $\frac{n^2}{n^2-n-5}$ converges I know the basic procedure for proving convergence of sequences, but I'm having difficulty reducing this sequence. I believe the sequence converges to 1, so I've set up the following two approaches:
1:
$\left|\frac{n^2}{n^2 - n - 5} - 1\right| = \left|\frac{n^2}{n^2 - n - 5} -
\frac{n^2 - n - 5}{n^2 - n - 5}\right| = \frac{n + 5}{n^2 - n - 5}$
2:
$\left|\frac{n^2}{n^2-n-5} - 1\right| = \left|\frac{1}{1-\frac{1}{n}-\frac{5}{n^2}} - \frac{1-\frac{1}{n}-\frac{5}{n^2}}{1-\frac{1}{n}-\frac{5}{n^2}}\right| = \frac{\frac{1}{n}+\frac{5}{n^2}}{1-\frac{1}{n}-\frac{5}{n^2}}$
Both assume $n \geq 3$. Neither of these approaches appear to be leading me in the right direction. I can technically solve these for $f(n) < f(\epsilon)$, but both result in something very nasty, and I feel like I'm missing something simple.

For $n\to \infty$ we have $\frac{n^2}{n^2-n-5}=\frac{1}{1-\frac{1}{n}-\frac{5}{n^2}}$ (dividing numerator and denominator by $n^2$). All terms on the denominator go to 0 except 1 so the limit is 1.
