Find $\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}})$

Question

$$\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}})$$ 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.

Answer 1

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 $1$?

Answer 2

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

Answer 3

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.

Answer 4

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?