I already determined that $\lim_{x\to0}x\sin\frac{1}{x}=0$ via the Squeeze Theorem, but I'm interested solving the limit without it. This is the work I have thus far:
$$\lim_{x\to0}x\sin{\frac{1}{x}}=\lim_{x\to0}x\times\lim_{x\to0}\sin\frac{1}{x}=\lim_{x\to0}x\times\sin(\lim_{x\to0}\frac{1}{x})=0\times\sin(\lim_{x\to0}\frac{1}{x})=0$$
My concerns with my work is that $\lim_{x\to0}\frac{1}{x}$ doesn't exist, which would technically make the entire limit undefined. I tried to use the fact $\lim_{x\to0}x=0$ to justify my answer, but it assumes $\times$ is evaluated before $\lim$. Therefore, my questions are as follows:
- Is $\times$ evaluated before $\lim$? If so, does that make my work correct?
- If not, how would I overcome the issue of $\lim_{x\to0}\frac{1}{x}$ to get a numerical answer to the limit without using the Squeeze Theorem?
EDIT: It appears I didn't clarify my question enough; I'm looking to solve the limit without the Squeeze Theorem
Best Answer
Hint
Since $|\sin y|\le 1, \forall y\in\mathbb{R},$ we have that
$$\left| x\sin\dfrac1x\right|\le \left| x\right|,$$ and, thus,
$$\left|\lim_{x\to 0}\left( x\sin\dfrac1x\right)\right|\le \left|\lim_{x\to 0} x\right|.$$