0
$\begingroup$

As a part of homework, I was asked

What does $\lim_{x\to a} f(x)=\infty$ mean?

In an earlier calculus class I was taught that in order for $L=\lim_{x\to a}f(x)$ to exist, we need that $L=\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)$. In an explicit(but extremely non-rigorous) example the teacher explained with a graphic in which the two two sides of a vertical asymptote tended to $\infty$, the limit didn't exist because we couldn't equate both lateral limits, since we cannot compare two infinities. Is this right? If so, I would answer "It doesn't make sense, since we would need that both lateral limits at $a$...equated to the same infinity?" If it is not right and that does make some sense, may you clarify this point to me about that definition?

$\endgroup$
  • 1
    $\begingroup$ In it's current form, it is meaningless. The limit of what? As for the definition, one can make sense of it as follows: $\forall M \in \mathbb{R}$, there exists $\delta > 0$ such that $0 < |x-a| < \delta$ (note my initial error, it should be a deleted neighbourhood) implies $f(x) > M$. The one-sided limit on the right would be $0 < x - a < \delta$; the one on the left would be $0 < a - x < \delta$. The two-sided limit only makes sense if they are equal. $\endgroup$ – Chris K Apr 2 '14 at 0:19
  • $\begingroup$ @ChrisK Sorry, I was kind of absentminded when I wrote this, the post is edited now. Using that $\epsilon-\delta$ definition, it is clear that both limits tend to $\infty$. But is it common to write that the limit is $\infty$, even if we can't equate both lateral lmimits? $\endgroup$ – chubakueno Apr 2 '14 at 0:29
  • $\begingroup$ If both limits aren't equal, we can only make sense of one-sided limits. We then say the two-sided limit does not exist. $\endgroup$ – Chris K Apr 2 '14 at 1:23
  • $\begingroup$ @chubakueno If it helps, think of $\infty$ as notation for "grows without bound", rather than representing some actual value. $\endgroup$ – augurar Apr 3 '14 at 3:39
1
$\begingroup$

$\lim_{x\to a}f(x)\to\infty$ means that as you go closer and closer to $a$, the value of $f(x)$ grows arbitrarily large. Now, if you approach $-\infty$ as you go closer to $a$ from the left (or right), and if you approach $+\infty$ as you go closer to $a$ from the right (or left), then the limit does not exist. The reason is that the limit has two values at the same point. To be mathematically precise, since we know that the left- and right-hand limits are the same if and only if the ordinary limit exists, and that the left- and right-hand limits are not the same, $\lim_{x\to a}f(x)$ is undefined. I believe that this is what you mean by "two sides of a vertical asymptote tended to $∞$."

$\endgroup$
  • $\begingroup$ Yes, that was what I meant. Alas(it seems) there is not an universal convention about this. The epsilon delta arguments can prove that a function can grow without bound towards $+\infty$, but with the comments and answers here I think that $\lim_{x\to a} f(x)=\infty$ is not universally accepted. $\endgroup$ – chubakueno Apr 2 '14 at 0:50
  • $\begingroup$ @chubakueno Yes, that's true... $\endgroup$ – user122283 Apr 2 '14 at 0:57
1
$\begingroup$

There is still some difference of feeling on how to talk about these. In the traditional discussion of limits, a limit should be a finite number that is "approached arbitrarily closely" by the function (made rigorous by $ \ \epsilon - \delta \ $ proofs, or the like). So when a function does not approach a finite value as $ \ x \ \rightarrow \ a \ , $ we say the limit "does not exist".

But there are various ways this can happen. The oxymoronic phrase "infinite limit" is introduced in some texts to indicate that a function grows "without limit" to the same signed infinity for $ \ x \ $ approaching $ \ a \ $ "from both sides". Hence, these authors will write $ \ \lim_{x \rightarrow 0} \ \frac{1}{x^2} \ = \ +\infty \ , $ for example; they would still write $ \ \lim_{x \rightarrow 0} \ \frac{1}{x} \ $ DNE .

$\endgroup$
  • 1
    $\begingroup$ I think you have a typo... $lim_{x->0} \frac{1}{x^2}$ is certainly not $0$! $\endgroup$ – augurar Apr 2 '14 at 0:24
  • $\begingroup$ Oops, sorry! Thinking too much about the zeroes. Thanks. $\endgroup$ – colormegone Apr 2 '14 at 0:25
  • $\begingroup$ So It means that it grows without limit to the positive infinity, even though lateral limits are not equal would be the most politically correct answer? $\endgroup$ – chubakueno Apr 2 '14 at 0:35
  • $\begingroup$ I'm not sure exactly what a "lateral limit" is; your post is the first time I've seen the term. How does it differ from a "one-sided" limit? The issue of "comparing infinities" is being avoided, since the notion of calling something an "infinite limit" is just a convention to summarize a particular situation (and one which is not universally accepted). $\endgroup$ – colormegone Apr 2 '14 at 0:44
  • $\begingroup$ We are talking about the same thing, it just just my bad use of english :) (that I am currently trying to improve) $\endgroup$ – chubakueno Apr 2 '14 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.