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$$\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}})$$ First I don't know how to solve this problem since there is $\sin (\frac{1}{\sqrt[3]{x-1}})$ in there. But I manage to do this: $$\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}}) = (\lim\limits_{x \to 1} (x-1)^2) (\lim\limits_{x \to 1}\sin (\frac{1}{\sqrt[3]{x-1}})) $$ $$= ((1-1)^2) (\lim\limits_{x \to 1}\sin (\frac{1}{\sqrt[3]{x-1}}))$$ $$= (0) (\lim\limits_{x \to 1}\sin (\frac{1}{\sqrt[3]{x-1}}))$$ $$= 0$$ But I'm pretty sure that I'm wrong here.

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    $\begingroup$ There is a problem in your first step. You can't do it because one of the limits does not exist. $\endgroup$ – Juniven Apr 2 '17 at 15:40
  • $\begingroup$ I know, that's why I ask for help here. $\endgroup$ – Difa Sanditya Apr 2 '17 at 15:47
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Well, according what I read. Using, the squeeze theorem meaning that:
$$-1 \le \sin (\frac{1}{\sqrt[3]{x-1}}) \le 1$$ $$-(x-1)^2 \le \sin (\frac{1}{\sqrt[3]{x-1}}) \le (x-1)^2$$ $$-(x-1)^2 \le (x-1)^2\sin (\frac{1}{\sqrt[3]{x-1}}) \le (x-1)^2$$ $$\lim\limits_{x \to 1} -(x-1)^2 \le \lim\limits_{x \to 1} (x-1)^2\sin (\frac{1}{\sqrt[3]{x-1}}) \le \lim\limits_{x \to 1} (x-1)^2$$ Then I got this: $$\lim\limits_{x \to 1} -(x-1)^2 = -(1-1)^2 = 0$$ And this: $$\lim\limits_{x \to 1} (x-1)^2 = (1-1)^2 = 0$$ Meaning that: $$0 \le \lim\limits_{x \to 1} (x-1)^2\sin (\frac{1}{\sqrt[3]{x-1}}) \le 0$$ So, I can conclude that: $$\lim\limits_{x \to 1} (x-1)^2\sin (\frac{1}{\sqrt[3]{x-1}}) = 0$$ Is this the correct way?

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  • $\begingroup$ Yes, you did it correctly. But you need to say in the first few steps that $x\neq 1$. $\endgroup$ – Juniven Apr 3 '17 at 4:48
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You're being sloppy in your final step, but if you squint and mumble the words "squeeze theorem", your answer is right in spirit. Can you use the squeeze theorem more formally, noting that the $\sin$ term is always between $-1$ and $1$?

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  • $\begingroup$ Does that mean I have to use $-1 \lt \sin (\frac{1}{\sqrt[3]{x-1}}) \lt 1$ ? $\endgroup$ – Difa Sanditya Apr 2 '17 at 15:36
  • $\begingroup$ I would use $\leq$ rather than $<$, but yes. Squeeze the original function between $\pm (x-1)^2$. $\endgroup$ – Patrick Stevens Apr 2 '17 at 15:36
  • $\begingroup$ Well, I think I need to understand more about this squeeze theorem, and will be right back as soon as I understand it. Thanks by the way. $\endgroup$ – Difa Sanditya Apr 2 '17 at 15:46
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For all $x\neq 1$, we have $$\begin{align} 0&\leq\bigg|(x-1)^2\sin\frac{1}{\sqrt[3]{x-1}}\bigg|\\ &=|(x-1)^2|\cdot \bigg|\sin\frac{1}{\sqrt[3]{x-1}}\bigg|\\ &\leq |(x-1)^2|\cdot 1\\ &=(x-1)^2 \end{align}$$ Because $$\lim_{x\to 1} 0=0=\lim_{x\to 1}(x-1)^2,$$ it follows from the Squeeze Theorem that $$\lim_{x\to 1}\bigg|(x-1)^2\sin\frac{1}{\sqrt[3]{x-1}}\bigg|=0$$ and hence, we get $$\lim_{x\to 1}(x-1)^2\sin\frac{1}{\sqrt[3]{x-1}}=0$$

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  • $\begingroup$ Well, I don't really understand about this squeeze theorem. But I think I found my own way, would you mind checking my answer? Thanks by the way. $\endgroup$ – Difa Sanditya Apr 3 '17 at 2:57
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You have a bounded expression times an expression $\to 0.$ The outcome is always the same in such a situation: The limit is $0.$ It is very important that you understand this concept.

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  • $\begingroup$ So basically my first answer is correct since anything times a zero equal to zero, right? $\endgroup$ – Difa Sanditya Apr 3 '17 at 2:58
  • $\begingroup$ No, it's incorrect because you have written $\lim_{x\to 1}\sin(1-x^3)^{-1/3},$ which makes no sense at all. $\endgroup$ – zhw. Apr 3 '17 at 23:00

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