calculus - The Squeeze Theorem

Find

$$\lim_{x\to0}\left(x^2\sin\frac1x\right)$$

Solution:

We have:

$$\color{red}{-1\le\sin\frac1x\le1}\;;\;x\in(-5,5)-\{0\}\quad\text{Why?}$$

(Multiplying by $x^2$ …Note that $x^2$ is non-negative)

$$\implies-x^2\le x^2\sin\frac1x\le x^2\;;\;x\in(-5,5)-\{0\}$$

We have

$$\lim_{x\to0}\left(-x^2\right)=0=\lim_{x\to0}\left(x^2\right)$$

$\implies$ By the squeeze theorem,

$$\lim_{x\to0}\left(x^2\sin\frac1x\right)=0$$

Can someone please explain this problem? I could solve the ones without $\sin$/$\cos$.

Why do we assume that $\sin\dfrac1x$ is between $-1$ and $1$?