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!
 A: Naively, when $n \to \infty$, $n^2+2$ is like $n^2$ and $2n^2-1$ is like $2n^2$, so the limit should be $\frac{n^2}{2n^2} = \frac{1}{2}$. To make this rigorous, the easiest is probably to divide by $n^2$ the top and bottom :
$$
\lim_{n\to \infty} \frac{n^2+2}{2n^2-1} = \lim_{n\to \infty} \frac{1+2/n^2}{2-1/n^2}
$$
Now we know that $1/n^2 \to 0$ when $n \to \infty$ so the usual limit rules show that the answer is indeed $1/2$.
A: You might find it easier to write
$$\dfrac{n^2+2}{2n^2-1} = \dfrac{1}{2}  \times \dfrac{2n^2+4}{2n^2-1} = \dfrac{1}{2} + \dfrac{5}{4n^2-2}$$ and note you can make $\dfrac{5}{4n^2-2}$ as small as you want by making $n$ large enough.
If you must use epsilon-delta or epsilson-$N_0$, then note that $\dfrac{5}{4n^2-2} \lt \varepsilon $ if $\dfrac5\varepsilon \lt 4n^2 -2$ which is true if $n \gt \sqrt{\dfrac12+\dfrac{5}{4\varepsilon}}$  
A: $\left|\frac{n^2+2}{2n^2-1} - \frac {1}{2}\right| =  \frac{5}{4n^2-2}$
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 $n\geq n_0$.
