Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I heard that $0/0$ is undetermined, but why is $\infty/\infty$ also undetermined? Also what is $\lim\limits_{x \to \infty} \frac{x}{x}$ and what is $\lim\limits_{x \to \infty} \frac{2^x}{2^x}$?

share|improve this question
2  
Well, contrast the limiting behaviors of $\exp(x)/x$ and $x/\exp(x)$ for instance... –  J. M. Jul 22 '11 at 18:42
    
@J.M.-what's so special about the contrasting? –  abcde Jul 22 '11 at 18:45
    
Graph both and observe what happens to them as you go rightward... –  J. M. Jul 22 '11 at 18:49
4  
Hmm. Didn't you learn just yesterday how to prove that one of them is zero? The same argument shows that the other one is infinity. Since you seem to like l'Hôpital, use that one! –  t.b. Jul 22 '11 at 18:51
2  
@abcde: Nothing wrong with using a calculator to get some numerical insight. –  André Nicolas Jul 23 '11 at 1:49

5 Answers 5

up vote 4 down vote accepted

A short and non rigorous explanation: in a fraction $\dfrac{P(x)}{Q(x)}$ (assume that both $P(x)$ and $Q(x)$ are positive) it may happen that $P(x)$ goes to infinity "faster", "slower" or "as fast as" $Q(x)$. Depending on these speeds the fraction tends to infinity, zero or a constant, or it may happen that the limit does not exist, as commented by Samuel. To find the limit one can apply L'Hôpital's rule, if $P(x)$ and $Q(x)$ are differentiable.

Both $\lim\limits_{x \to \infty} \frac{x}{x}$ and $\lim\limits_{x \to \infty} \frac{2^x}{2^x}$ is $1$.

share|improve this answer
    
To add to this: It may also happen that the limit does not exist. –  Samuel Jul 23 '11 at 21:22
    
@Samuel: That's right. I will add this possibility to the answer. –  Américo Tavares Jul 23 '11 at 21:42

Your question is completely reasonable, given the unfortunate survival of inappropriate notation that has confused many generations of students.

Here I am referring to phrases of the kind "indeterminate form of the type $\infty/\infty$," which one sees even in very good calculus books.

This notation encourages the notion that there is such a thing as the "number" infinity, that it makes sense to divide this "number" by things, including "$\infty$," but that somehow the result of this so-called division can be various things. Sadly, we also meet "$\cdot \infty$," and "$\infty-\infty$."

After a while, most students either find out what's really going on, or else at least understand that they must, for some unknown reason, use special rules, and manage to get by.

The reality is really quite simple. We have two functions, $f(x)$ and $g(x)$ such that when $x$ gets very large, each of $f(x)$ and $g(x)$ both get very large. The question is: What happens to $$\frac{f(x)}{g(x)}$$ as $x$ gets very large?

The answer is: It depends. Here are some examples.

1. Let $f(x)=x^2+17x$, and $g(x)=x^3+1$. Each gets very large when $x$ gets large. But it is fairly easy to see that after a while, $x^3+1$ is much larger than $x^2+17x$, so after a while the ratio is close to $0$.

2. Let $f(x)=x^2+17x$, and $g(x)=5x^2 -11$. When $x$ is large, the behaviour of $f(x)$ is dominated by the $x^2$ term, and the behaviour of $g(x)$ is dominated by the $5x^2$ term. So the ratio $f(x)/g(x)$ should be more or less $x^2/5x^2$ when $x$ is large. One can informally verify with a calculator that it indeed looks as if $f(x)/g(x)$ is close to $1/5$ when $x$ is large.

3. Let $f(x)=e^x$, and $g(x)=x^{10}$. Here things are not obvious. By remembering that, in the long run, $e^x$ grows faster than any polynomial, we can expect that $f(x)/g(x)$ gets very large as $x$ gets large. But maybe in this case we should use some reliable machinery to find out what happens.

4. There can be more complicated behaviour, with $f(x)$ being way ahead for a while, then falling way behind, then racing ahead, and so on. It takes some work to come up with examples of this kind of behaviour, but it can happen.

No "$\infty/\infty$" anywhere! All we have been looking at is questions about the long-term behaviour of certain ratios of functions. Fairly often, we need information about such ratios. In many cases, intuition is not sufficiently accurate to provide an answer, so we need special tools. In the case of your two questions, we need no tools, since in each case the ratio is $1$ for all $x$.

The term $\infty/\infty$ should be viewed merely as a label that calculus books to questions of the types discussed above. It is a terribly unfortunate label, since it invites misunderstanding.

One can hope that the label $\infty/\infty$, and its brethren, such as $0/0$, will someday disappear from calculus books. But we should not hold our breath: they have been around since the early eighteenth century!

share|improve this answer

HINT $\rm\ \ f/g\ $ of form $\:0/0$ $\rm\ \Rightarrow\ (f/g)^2 =\: (1/g^2)/(1/f^{\:2})\ $ of form $\:\infty/\infty$

share|improve this answer

Actually, $\frac{\infty}{\infty } = \frac{5}{2}$. Don't believe me? Look. Take these three facts:
(1) $\lim_{x\to\infty} 5x+2 = \infty$
(2) $\lim_{x\to\infty} 2x = \infty$
(3) $\lim_{x\to\infty} \frac{5x+2}{2x} = \frac{5}{2}$
Now, divide (1) by (2) and compare with (3) to conclude $\frac{\infty}{\infty } = \frac{5}{2}$.

share|improve this answer

Your question has some issues, since it depends on what you mean by $\frac 0 0$ and $\frac \infty \infty$.

Either you're asking why $\frac 0 0$ is not defined and why $\frac \infty \infty$ is not defined. The latter one would be a meaningless statement in $\mathbf R$ since $\infty$ does not denote a real number. Generally it is either an indicator that a limit diverges (in which case an equality statement is something of an abuse of notation) or an element of an extension of $\mathbf R$ where $\infty$ is an "added" element greater than any other. There are some issues with doing calculus in this structure though (and I would hazard a guess that this is not what you are trying to do).

The former expression being meaningless is due to $0$ being identity of addition of what is known as a field. Quickly the reason for this is that division $\frac a b $ is then defined as $a \cdot b^{-1}$, where $b^{-1}$ is the (unique) element ssuch that $b \cdot b^{-1} = 1$, but clearly there cannot exist such a $0^{-1}$ since $0 \cdot a = 0 \not = 1$ (this is generally true for fields, and especially for $\mathbf R$) for any $a$.

My other interpretation is why there is no general value for $\lim \limits_{x \to a} \frac {f (x)} {g (x)}$ for $f, g$ such that either (corresponding to the first statement) $\lim \limits_{x \to a} \; f(x) = 0$ and $\lim \limits_{x \to a} \; g(x) = 0$ or (corresponding to the second statment) $\lim \limits_{x \to a} \; f(x) = \infty$ and $\lim \limits_{x \to a} \; g(x) = \infty$. In this case it's just as the other answers have suggested, that different functions can tend differently "fast" to their limits (or their lack thereof).

As for your example functions $\frac x x$ and $\frac {2^x} {2^x}$ both are equal to the constant function $f(x) = 1$, so their limits (when x tends to any value) is just 1.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.