Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials

Short Answer:

The sequence $\displaystyle\frac{Q(k)}{P(k)}$ will converge to the same limit as the function $\displaystyle\frac{Q(x)}{P(x)}.$ There are three cases:

$(i)$ If $n>m$ then it diverges to either $\infty$ or $-\infty$ depending on the sign of $\frac{a_{n}}{b_{m}}$.

$(ii)$ If $n<m$ then it converges to $0$.

$(iii)$ If $n=m$ then it converges to $\frac{a_{n}}{b_{n}}$.

More generally: if a sequence $a_n$ is given by the values of function $f(x)$ that is defined on an interval of the form $(b,\infty)$, $$a_n = f(n),$$ and the limit of $f(x)$ as $x\to\infty$ exists or is equal to $\infty$ or $-\infty$, $$\lim_{x\to\infty}f(x) = L,\qquad L\in\mathbb{R}\cup\{\infty,-\infty\},$$ then the limit of the sequence is the same as the limit of the function: $$\lim_{n\to\infty}a_n = \lim_{n\to\infty}f(n) = \lim_{x\to\infty}f(x).$$

This applies to the case where $\displaystyle f(x)= \frac{P(x)}{Q(x)}$ with $P$ and $Q$ polynomials; also to sequences like $$a_n = \frac{\sin(n)}{n},$$ given by $\displaystyle f(x) = \frac{\sin(x)}{x}$; and even some functions which are more complicated. E.g., $$a_n = \frac{(-1)^n}{n}$$ can be seen as given by the function $$f(x) = \frac{\cos(\pi x)}{x}.$$

Note, however, that it is possible for the limit of $a_n$ to exist, but that of $f(x)$ not to exist. For instance, the sequence $a_n = \sin(n\pi)$ has limit $0$ (because every $a_n$ is equal to $0$), but the limit of $f(x)=\sin(\pi x)$ as $x\to\infty$ does not exist.