show $\lim_{x\to 0}x\sin\frac{1}{x}=0$ without the Squeeze Theorem 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
 A: 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,
$$\left|\lim_{x\to 0}\left( x\sin\dfrac1x\right)\right|\le \left|\lim_{x\to 0} x\right|.$$
A: Your fundamental problem is the use of product rule of limits. The limit of $\sin(1/x)$ does not exist and unfortunately the limit of other factor is $0$. Note that the product rule can be applied if one of the factors has a non-zero limit. 
So that reasoning does not work. Next you can notice that $|x\sin(1/x)|\leq |x|$ and hence you may guess that the desired limit is $0$ and formally prove this using definition of limit. More generally the definition of limit can be used to prove the following simple result:

Theorem: If $f(x) \to 0$ as $x\to a$ and $g(x) $ is bounded as $x\to a$ then $f(x) g(x) \to 0$ as $x\to a$. 

A: Using squeeze theorem!
We know that The value of x can take a maximum of 1 and a minimum of negative one!
$$-1 \leq \sin x \leq 1$$
So the same applies to 
$$-1 \leq \sin \frac{1}{x} \leq 1$$
Now let's multiply the inequality with $x$,
$$-x \leq x\sin \frac{1}{x} \leq x$$
Taking the limit for all three quantities!
$$\lim_{\text x \rightarrow 0}-x \leq \lim_{\text x \rightarrow 0} x\sin \frac{1}{x} \leq \lim_{\text x \rightarrow 0} x $$
Know that 
$$\lim_{\text x \rightarrow 0}-x =0 $$
$$\lim_{\text x \rightarrow 0} x=0$$
So limit for $x \sin \frac{1}{x}$ is 
$$\lim_{\text x \rightarrow 0} x\sin \frac{1}{x}=0$$
Another way of seeing this
$$\lim_{x \rightarrow 0}x\sin\frac{1}{x}=\lim_{x \rightarrow 0}\frac{sin\frac{1}{x}}{\frac{1}{x}}$$
let $k=\frac{1}{x}$
$${x}\rightarrow0,k\rightarrow \infty$$
$$\lim_{k \rightarrow \infty}\frac{\sin k}{k}=0$$
A: this is not correct, use the squeeze Theorem:
$$\left|x\sin\left(\frac{1}{x}\right)\right|\le |x|$$
