To prove limit of function

Question

To prove $\lim_{x\to 0} x\sin(\frac{1}{x})=0$ I tried sandwich theorem but I have no clear idea to prove the problem

Answer 1

The sine of every real number is between $-1$ and $1$. Therefore $x \cdot \sin(1/x)$ is between $x\cdot(\pm1) = \pm x.$ And $+x$ and $-x$ both approach the same limit as $x\to0$.

Answer 2

Hint: $|\sin(x)|\leq 1 \quad \quad$

Answer 3

It can be sandwiched between $x$ and $-x$ (strictly speaking $|x|$ and $-|x|$).

Answer 4

$-|x| \leq x\sin(\frac{1}{x}) \leq |x| $. Hence

$\lim_{x \rightarrow 0} -|x| = 0 \leq \lim_{x \rightarrow 0} x\sin(\frac{1}{x})\leq \lim_{x \rightarrow 0}|x| =0 $.

Therefore,$ \lim_{x \rightarrow 0}x\sin(\frac{1}{x})=0 $ by squeezed theorem.