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.