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Suppose that $Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} $and $P(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}.$ How do I find $\lim_{x\rightarrow\infty}\frac{Q(x)}{P(x)}$ and what does the sequence $\frac{Q(k)}{P(k)}$ converge to?

For example, how would I find what the sequence $\frac{8k^2+2k-100}{3k^2+2k+1}$ converges to? Or what is $\lim_{x\rightarrow\infty}\frac{3x+5}{-2x+9}?$

This is being asked in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.

and here: List of abstract duplicates.

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    @Bill: To answer your questions: I prefer it when a new question is created and devoted solely to the abstract duplicate, rather than just hijacking one that was already asked. I think it makes things nicer and cleaner, however that is a matter of personal taste. As for the community wiki, I think that is a good idea, and have flagged the post for moderator attention. At first glance, I just went with what the other answers on the abstract duplicate page had done. (Upon a quick look again, only one is community wiki, but then again I think most were modifications of other questions)2011-04-20

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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 then it converges to $0$.

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

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    Lang had an interesting approach: factor $x^{n}$ out of the numerator and $x^{m}$ from the denominator to get $x^{n-m}$ times a rational function with constant limit $a_{n}/b_{m}$ as $x \rightarrow \infty$. The result Eric quoted now follows by examining the limit of $x^{n-m}$.2011-04-20
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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.