# How can $\mathop{\lim}\limits_{x \to \infty}({x+x\space\sin^2 x})$ exist if $\mathop{\lim}\limits_{x \to \infty}(x\space\sin^2 x)$ does not?

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?

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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. – raxacoricofallapatorius Sep 20 '11 at 23:34
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 $\infty$. – raxacoricofallapatorius Sep 20 '11 at 23:34
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, $\lim x + \sin x = \infty$. – TMM Sep 20 '11 at 23:54
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 $\mathbb{R}\cup\{\infty,-\infty\}$. – Brian M. Scott Sep 20 '11 at 23:54
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.) – 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 $x > a$.

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Would it have been correct to say $f(x)$ is sometimes $\infty$? – raxacoricofallapatorius Sep 21 '11 at 1:06
@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. – TMM Sep 21 '11 at 1:13
@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.) – raxacoricofallapatorius Sep 21 '11 at 1:25
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]? – raxacoricofallapatorius Sep 21 '11 at 1:43
@raxacoricofallapatorius, I just saw your question. I would assume [A] requires that the limit exist and be finite. – Hans Parshall Feb 1 '12 at 16:41