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

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.

• Bad form to use $n$ for both the degree of $Q$ and the integer argument of $Q$ in the same problem (but I think we all know what you mean). Apr 20, 2011 at 2:07
• @Gerry Myerson: Good point. Fixed now! Apr 20, 2011 at 2:09
• Probably also good to leave a note that the techniques here are useful for finding asymptotes of rational functions. Apr 20, 2011 at 2:10
• Eric: Such questions and answers should be CW in my opinion. Further, why not simply modify one of the many prior answers on such topics, so to reduce the noise? Apr 20, 2011 at 2:14
• @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) Apr 20, 2011 at 3:37

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 $\frac{a_{n}}{b_{n}}$.

• And perhaps the easiest way to get to this answer is to divide top and bottom by $x^d$, where $d$ is the smaller of $m$ and $n$. Apr 20, 2011 at 2:06
• 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 $x^{n-m}$. Apr 20, 2011 at 15:27

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.