In the text I'm using (Spivak's Calculus, 4E), it is established (problem 5.39(iii)) that $$\mathop{\lim}\limits_{x \to \infty}\left({x\space\sin^{2} x}\right)$$ "does not exist". It is also established (5.39(c)) that [A] if $\mathop{\lim}\limits_{x \to \infty}f(x)$ exists, but $\mathop{\lim}\limits_{x \to \infty}g(x)$ does not, then $\mathop{\lim}\limits_{x \to \infty}\left[f(x)+g(x))\right]$ cannot exist.

But the text also establishes (5.39(ii)) that $$\mathop{\lim}\limits_{x \to \infty}\left({x+x\space\sin^{2}x}\right)=\infty.$$

which seems to be a contradiction of the just established property of limits, with $f(x)=x$ and $g(x)=x\space\sin^{2} x$.

Is there something important going on here with regard to limits that "equal" infinity?

  • 1
    $\begingroup$ Looking closely at the definition of the limit, it would seem not to apply to $L=\pm\infty$, which suggests that saying that $\mathop{\lim}\limits_{x \to \infty}f(x)=\infty$ is just a particular way of saying that though the limit does not exist, we know that that the function grows without bound. $\endgroup$ – orome Sep 20 '11 at 23:34
  • $\begingroup$ If that (or something like it) is that case, I can see that while $\mathop{\lim}\limits_{x \to \infty}f(x)$ does not exist because it is only sometimes $\infty$, this shorthand can be used for $\mathop{\lim}\limits_{x \to \infty}\left[f(x)+g(x)\right]$ because, though it never settles on a single value, all the values it takes on for $x\to\infty$ are $\infty$. $\endgroup$ – orome Sep 20 '11 at 23:34
  • $\begingroup$ Indeed, by "adding $\infty$" the limit suddenly does exist, as then the oscillation becomes "irrelevant". A similar example is $f(x) = x$ and $g(x) = \sin x$. While $\lim \sin x$ does not exist, $\lim x + \sin x = \infty$. $\endgroup$ – TMM Sep 20 '11 at 23:54
  • 1
    $\begingroup$ You’ve answered your own question: it’s a matter of slightly sloppy terminology. The limit does not exist in this context means there is no extended real number that is the limit, where the extended reals are $\mathbb{R}\cup\{\infty,-\infty\}$. $\endgroup$ – Brian M. Scott Sep 20 '11 at 23:54
  • 1
    $\begingroup$ I always find it helpful to remember that the string of symbols $\lim f(x) = \infty$ does not mean that the limit exists; rather, it gives information about the specific way in which the limit fails to exist. (The savvy can also consider the extended real numbers as @BrianM.Scott did, but it's easy to mislead oneself that way.) $\endgroup$ – Greg Martin Sep 21 '11 at 2:03

This does not contradict the established property of limits. If you look closely, those are established only when the limits exist! This provides a loophole for this case.

As for this limit, it goes back to what we mean when we write $\lim_{x \rightarrow \infty} f(x) = \infty$. In particular, this means for any large $M$ I choose, you can find some large enough $a$ such that $f(x) > M$ for all $x > a$.

For $x\sin^2(x)$, there are infinitely many positive values for $x$ that make $x\sin^2(x)$ zero. In particular, there exists some $M$ ($M = 0$ works) such that for all $a$, there exists some $b > a$ with $b\sin^2(b) = 0$. Convince yourself that this is the negation of the above definition, so that we've genuinely shown that $\lim_{x \rightarrow \infty} x\sin^2(x) \neq \infty$. Using a similar argument, I bet you can show that $\lim_{x \rightarrow \infty} x\sin^2(x)$ does not equal anything else, either. You're right above when you say "it never settles on a single value". However, you're incorrect when you say it is "sometimes $\infty$"; this makes no sense to say.

It should not be difficult to use the definition to show $\lim_{x \rightarrow \infty}(x + x\sin^2x) = \infty$. In particular, choose some arbitrary $M$ and find some value $a$ (it will depend on $M$!) such that $x + x\sin^2x > M$ for all $x > a$.

  • $\begingroup$ Would it have been correct to say $f(x)$ is sometimes $\infty$? $\endgroup$ – orome Sep 21 '11 at 1:06
  • $\begingroup$ @rax: $f(x)$ is never $\infty$, as for any $x$ I can give you an $M$ such that $|f(x)| < M$. Only when you turn the game around, and you tell me to first give an $M$ (and you then give $x$), you can show that $f$ is "unbounded". But then you're talking about limits. $\endgroup$ – TMM Sep 21 '11 at 1:13
  • $\begingroup$ @Thijs: Ah, I see (I think): For example, you could just give me $M=\lvert f(x) \rvert+1$ in response to my $x$; but (for $f(x)=x$) in response to your $M$, I could give $x=M +1$. (Also note that I mixed up my $f$s and $g$s in the "sometimes" comment we're referring to.) $\endgroup$ – orome Sep 21 '11 at 1:25
  • $\begingroup$ The "established property of limits" I was referring to was [A] in the question, which does refer to one function for which a limit does not exist. It also, however requires that another function does have a limit. So perhaps that's the crux of my question: does $f(x)=x$ have a limit in the sense required for [A]? $\endgroup$ – orome Sep 21 '11 at 1:43
  • $\begingroup$ @raxacoricofallapatorius, I just saw your question. I would assume [A] requires that the limit exist and be finite. $\endgroup$ – Hans Parshall Feb 1 '12 at 16:41

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.