Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$ 
Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$ for $n \to \infty$. 

I've been looking at this for hours! Also, sorry I don't have the proper notation. 
This is where I'm at:
$$
\left| \frac{n^2 + 2}{2 \cdot n^2 - 1} - \frac{1}{2}\right| = \left| \frac{5}{4 \cdot n^2 - 2} \right|
$$
I thought I was supposed to get to the point where I can say $1/n <n$, but I can only get to $1/n-1$ so that can't be the right approach or I'm missing something.
A friend says to make $n > \frac{5}{\varepsilon^2}$ but I''m not sure what to do with that tip. 
Any help would be greatly appreciated!

$\left|\frac{n^2+2}{2n^2-1} - \frac {1}{2}\right| =  \frac{5}{4n^2-2}$
Let $\epsilon>0$ be given. Let $n_0$ be the smallest integer such that $n\geq n_0> \sqrt{\frac{5}{4\epsilon} + \frac 12}$. Equivalently, $\epsilon>\frac{5}{4n^2-2}$.
Thus, $\left|\frac{n^2+2}{2n^2-1} - \frac {1}{2}\right|< \epsilon$, for all $n\geq n_0$.
