<s>[INST] Find $\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}})$ $$\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}})$$
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.
 [/INST] 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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