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?

Solutions Collecting From Web of "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?"

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 $x > a$.