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.
 A: 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. 
A: 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 $> n$.
A: 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 $n>N$
$$ \left|\frac{n+2}{3n^2-1}\right|< \frac{1}{n-2} < \epsilon \,. $$
A: We can use long polynomial division on this fraction to reduce it to terms which all "obviously" converge to zero.  I don't know how to typeset long division in Stackexchange TeX markup, but it only has a few steps in this case which can be described easily:
We start by strategically guessing the first term as $1\over 3n$. Then we multiply this by $3n^2 - 1$ to obtain the partial product $n - {1\over 3n}$.  To continue the long division we subtract $(n + 2) - (n - {1\over 3n})$ which gives us the remainder $2 + {1\over 3n}$.
At this point we can stop, and express our fraction as a sum of the term, plus the remainder divided by the divisor. In other words:
$${1\over 3n} + {{2 + {1\over 3n}\over 3n^2 - 1}}$$.
Now we can massage this further to separate out terms:
$${1\over 3n} + {{2 + {1\over 3n}\over 3n^2 - 1}}$$
$${1\over 3n} + {{2\over 3n^2 - 1}} + {{{1\over 3n}\over 3n^2 - 1}}$$
$${1\over 3n} + {{2\over 3n^2 - 1}} + {{{1\over 3n(3n^2 - 1)}}}$$
$${1\over 3n} + {{2\over 3n^2 - 1}} + {{{1\over 9n^3 - 3n}}}$$
As  you can see, your original fraction of two polynomials is a sum of three fractions, each of an integer divided by a polynomial.
In general, whenever you have a fraction of two polynomials $P(x)\over Q(x)$, then if $P$ has a lower degree than $Q$, the fraction has a limit of zero as $x$ tends to infinity. This is because $Q$ grows faster than $P$. If the highest degree terms in $P$ and $Q$ have some coefficients other than $1$ on them, that doesn't matter.
If $P$ and $Q$ have the same degree, then the limit will be the ratio of the coefficients of their degree terms. For example the limit of ${3x^2 + 5}\over 4x^2 - 16x + 9$ will just be $3/4$. As $x$ gets large, the $5$ term and the $-16x + 9$ terms become less and less significant, and the fraction more and more closely approaches ${3x^2}\over 4x^2$.
If $P$ has a larger degree than $Q$, then the fraction will not converge to zero. However, as $x$ grows large, that fraction will approximate a simple polynomial, which can be a very useful result. To find that polynomial, use long division to divide $P$ by $Q$. The quotient will be a simple polynomial, plus a fractional remainder. But the fractional remainder tends to zero as $x$ gets large, and can be ignored.
