Squeeze Theorem x squared times sin 1 over x

${\lim }_{x\to 0}{x}^2\sin (\frac{1}{x})$

Solution

Solving with squeeze theorem

1. Note that $\forall x\in R$, $-1\le \sin (x)\le 1$ defn of sine
2. Then $\forall x\in R$, $-1\le \sin (\frac{1}{x})\le 1$ choice of x
3. Then ${\lim }_{x\to 0}-x^2\le {\lim }_{x\to 0}{x}^2\sin (\frac{1}{x})\le {\lim }_{x\to 0}{x}^2$ As $x^2$ is positive
4. Then $0\le {\lim }_{x\to 0}{x}^2\sin (\frac{1}{x})\le 0$ limit properties
5. Thus ${\lim }_{x\to 0}{x}^2\sin (\frac{1}{x})=0$ by squeeze theorem