#### Answer

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x|\lt\delta\Rightarrow|f(x)|\lt\epsilon$$

#### Work Step by Step

$$\lim_{x\to0}x^2\sin\frac{1}{x}=0$$
Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x|\lt\delta\Rightarrow|f(x)|\lt\epsilon$$
1) We know that for all $x$, $$|\sin\frac{1}{x}|\le1$$
Therefore, $$|x^2\sin\frac{1}{x}|\le|x^2|$$
2) Now let an arbitrary value of $\epsilon\gt0$, and let a value of $\delta$ so that $\delta=\min(1,\epsilon)$.
There are only 2 cases:
- If $\epsilon\le1$, then $\delta=\epsilon\le1$, so $\delta^2\le\delta=\epsilon$.
- If $\epsilon\gt1$, then $\delta=1\lt\epsilon$, so $\delta^2=1\lt\epsilon$
Therefore, $\delta^2\le\epsilon$
For all $x$, as $0\lt|x|\lt\delta$, we have $$|f(x)|=|x^2\sin\frac{1}{x}|\le|x^2|\lt\delta^2$$
$$|f(x)|\lt\delta^2$$
Yet since $\delta^2\le\epsilon$, $$|f(x)|\lt\epsilon$$
The proof has been completed.