Answer
$$0=f^{\prime}(0)$$
Work Step by Step
Recall that :
\[
\begin{array}{c}
\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} =f^{\prime}(a)\\
\lim _{x \rightarrow 0} \frac{x^{2} \sin \left(\frac{1}{x}\right)-0}{x-0}=f^{\prime}(0) \\
\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=f^{\prime}(0)
\end{array}
\]
If
\[
h(x) \leq f(x) \leq g(x)
\]
,
\[
L=\lim _{x \rightarrow a} h(x)
\]
,
\[
L=\lim _{x \rightarrow a} g(x)
\]
Then
\[
L=\lim _{x \rightarrow a} f(x)
\]
since $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$
Thus$-x \leq x \sin \left(\frac{1}{x}\right) \leq x$
since
\[
0=\lim _{x \rightarrow 0}-x
\]
,
\[
0=\lim _{x \rightarrow 0} x
\]
By Squeeze Theorem, We have
\[
\begin{array}{c}
\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=0 \\
\end{array}
\]
