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.

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
$$\lim_{x\to 1}(x-1)^2\sin\frac{1}{\sqrt[3]{x-1}}=0$$
