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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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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). –  Gerry Myerson Apr 20 '11 at 2:07
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@Gerry Myerson: Good point. Fixed now! –  Eric Naslund Apr 20 '11 at 2:09
    
Probably also good to leave a note that the techniques here are useful for finding asymptotes of rational functions. –  J. M. Apr 20 '11 at 2:10
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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? –  Bill Dubuque Apr 20 '11 at 2:14
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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) –  Eric Naslund Apr 20 '11 at 3:37

2 Answers 2

up vote 6 down vote accepted

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

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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 $n$. –  Gerry Myerson Apr 20 '11 at 2:06
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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 $x^{n-m}$. –  Chris Leary Apr 20 '11 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.

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