# Prove/disprove: If $\lim_{x\to\infty}f(x)=\infty$, then $\lim_{x\to\infty}\sin(f(x))$ does not exist; and related question

Let $$f$$ be a function with domain $$\mathbb{R}$$. Is each of the following claims true or false? If it is false, show it with a counterexample. If it is true, prove it directly from the formal definitions of a limit.

• (a) IF $$\lim _{x\to\infty} f(x)=\infty$$, THEN $$\lim _{x\to\infty} \sin (f(x))$$ does not exist.
• (b) IF $$f(-1)=0$$ and $$f(1)=2$$, THEN $$\lim _{x\to\infty} f(\sin (x))$$ does not exist.

Original question image

I think the (a) is false and the (b) is true just through intuition, but I can't seem to come up with an example for the first one or understand how to prove the second one using the definitions.

• First one, simply let $f(x) =x$. Or any function that increases without bound. – Andrew Chin Oct 8 at 23:46
• @AndrewChin if we want to prove it to be true then a) f needs to be arbitrary and we can't pick a function and b) we need to prove directly from the limits. I feel as though there is a counterexample for the first one but I just can't seem to figure it out – adil.a Oct 8 at 23:48
• I am claiming that statement a) is false and am showing with a counterexample. – Andrew Chin Oct 8 at 23:49
• @AndrewChin did you misread the statement? – qbert Oct 8 at 23:55
• Let $f(x) = \lfloor x\rfloor *2\pi$. Then $\lim f(x) \to \infty$ and $\lim \sin f(x) = 0$. – fleablood Oct 9 at 0:01

Let $$f$$ take the constant value $$n\pi$$ on $$(n,n+1]$$ for each $$n$$. Then $$f(x) \to \infty$$ as $$x \to \infty$$ but $$f(\sin x )\equiv 0$$. So a) is false.
b) is true. Consider the points $$x=\pi /2 +2n\pi$$ and $$x=-\pi /2 +2n\pi$$ to see that $$f(\sin x )$$ cannot have a limit at $$\infty$$.
[ If possible let $$f(\sin x) \to l$$. Then there exists $$T$$ such that $$|f(\sin x)-l| <\frac 1 2$$ for all $$x >T$$. Put $$x=\pi /2 +2n\pi$$ (with $$n$$ large enough to make $$x >T$$) to see that $$|2-l| <\frac 1 2$$ and put $$x=-\pi /2 +2n\pi$$ (with $$n$$ large enough to make $$x >T$$) to see that $$|0-l| <\frac 1 2$$. It follows that $$|2-0| \leq |2-l| +|l-0| <\frac 1 2+\frac 1 2=1$$ which is a contradiction.