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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.

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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.

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  • $\begingroup$ First one, simply let $f(x) =x$. Or any function that increases without bound. $\endgroup$ – Andrew Chin Oct 8 at 23:46
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    $\begingroup$ @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 $\endgroup$ – adil.a Oct 8 at 23:48
  • $\begingroup$ I am claiming that statement a) is false and am showing with a counterexample. $\endgroup$ – Andrew Chin Oct 8 at 23:49
  • $\begingroup$ @AndrewChin did you misread the statement? $\endgroup$ – qbert Oct 8 at 23:55
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    $\begingroup$ Let $f(x) = \lfloor x\rfloor *2\pi$. Then $\lim f(x) \to \infty$ and $\lim \sin f(x) = 0$. $\endgroup$ – fleablood Oct 9 at 0:01
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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.

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  • $\begingroup$ Thank you for the response. How would you put part b in terms of epsilon and delta? $\endgroup$ – adil.a Oct 9 at 0:25
  • $\begingroup$ @adil.a I have added some details. $\endgroup$ – Kabo Murphy Oct 9 at 0:30

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