Prove limit of $\lim\limits_{x\mapsto 0}x\cdot\sin\dfrac1x = 0$?

Question

How do you prove

$$\lim\limits_{x\mapsto 0}x\cdot\sin\dfrac1x = 0$$

with limit definition and without limit definition?

My problem is that $\lim\limits_{x\mapsto 0}\sin\dfrac1x$ has not limit in zero so I can't use $$\lim\limits_{x\mapsto 0}(f(x)+g(x))=\lim\limits_{x\mapsto 0}f(x) + \lim\limits_{x\mapsto 0}g(x).$$

Answer 1

$$0\le \left| x\sin { \frac { 1 }{ x } } \right| \le \left| x \right| $$ then apply "squeeze theorem"

Answer 2

Since $-1\leq \sin(1/x)\leq 1$ we get $-|x|\leq x\sin(1/x)\leq |x|$ for $x\neq 0$.

Answer 3

we have

$\forall x\in \mathbb R^*$

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

and

$$|xsin(\frac{1}{x})|\leq |x|$$

or

$\forall x\in \mathbb R^*$

$-|x|\leq xsin(\frac{1}{x})\leq|x|$

thus we conclude that

$\lim_{x\to 0}xsin(\frac{1}{x})=0$

Answer 4

Apply the substitution $u = 1/x$ and let $u \to \infty$. It's the same limit.

Answer 5

We know that $\lim_{x\to 0}x=0$ and $\sin(1/x)$ is bounded. Combine this two we get the result.