Prove that the sequence $(n+2)/(3n^2 - 1)$ converges to the limit $0$

I need to prove that $$\lim_{n \to \infty} \frac{n+2}{3n^2 - 1} = 0.$$

I usually have no trouble with these types of proofs but the $-1$ in the denominator and the $n+2$ are making things a little more difficult than usual. Can someone start me off in the right direction? (This is a for all $n \ge N(\varepsilon)$ type proof)

Thanks.

Hint For $n\ge 2$, we have $n+2\le 2n$ and $3n^2-1\gt 2n^2$. So our fraction is less than $\dots$.

Remark: For problems of this type, you know what the answer will be, since the top is negligibly small compared to the bottom. But the problem is that $n+2$ is a little bigger than $n$, and $3n^2-1$ is a little smaller than $3n^2$. Both of these inequalities are running the wrong way. However, we can alter both of these without changing the essence of the "top much smaller than bottom." It is often useful to know what can be safely given away.

What do you think about simplifying it first?

$$\dfrac{n+2}{2n^2-1} = \dfrac{1 + \frac{2}{n}}{2n - \frac{1}{n}}$$

So for $n > 2$, the numerator $< 2$ and the denominator $> n$.

Given $\epsilon >0\,,$ we must find $N$ such that $$ n > N \implies \left| \frac{n+2}{3n^2 - 1} - 0 \right| < \epsilon \,.$$

Notice that

$$\left|\frac{n+2}{3n^2-1}\right| < \left|\frac{n+2}{n^2-4}\right|=\left|\frac{1}{n-2}\right| =\frac{1}{n-2} < \epsilon \,. $$

$$ \implies n - 2 > \frac{1}{\epsilon} \implies n > \frac{1}{\epsilon} + 2 > \frac{1}{\epsilon}\,. $$

Therefore if we choose $N>\frac{1}{\epsilon}\,,$ for any $n>N$

$$ \left|\frac{n+2}{3n^2-1}\right|< \frac{1}{n-2} < \epsilon \,. $$