Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 52

Determine whether $f'(0)$ exists in the function
$
\displaystyle
f(x) = \left\{
\begin{array}{c}
x^2 \sin\left(\frac{1}{x}\right) & \text{if} & x \neq 0\\
0 & \text{if} & x = 0
\end{array}\right.
$



Based from the definition,



$
\displaystyle
f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x-a}
$



$
\begin{equation}
\begin{aligned}
f'(0) & = \lim\limits_{x \to 0} \frac{x^2 \sin \left( \frac{1}{x}\right) - f(0)}{x-0}\\
f'(0) & = \lim\limits_{x \to 0} \frac{x\cancel{^2}\sin \left( \frac{1}{x}\right)}{\cancel{x}}\\
f'(0) & = \lim\limits_{x \to 0} x \sin \left(\frac{1}{x}\right)
\end{aligned}
\end{equation}
$



Note that we cannot use $\displaystyle \lim\limits_{x \to 0} x \sin \left(\frac{1}{x}\right) =
\lim\limits_{x \to 0} x \cdot \lim\limits_{x \to 0} \sin \left(\frac{1}{x}\right)$



because $\displaystyle \lim\limits_{x \to 0} \sin \left(\frac{1}{x}\right)$ does not exist. However, since



$\quad\displaystyle -1 \leq \sin \left(\frac{1}{x}\right) \leq 1$



We have,



$\quad\displaystyle -x^2 \leq \sin \left(\frac{1}{x}\right) \leq x^2$



We know that,



$\quad\displaystyle \lim\limits_{x \to 0^-} (-x^2) = -0 = 0 \quad \text{ and } \quad \lim\limits_{x \to 0^+} x^2 = 0$



Taking $f(x) = -x^2$, $\displaystyle g(x) = x^2 \sin \left(\frac{1}{x}\right)$ and $h(x) = x^2$ in the squeeze theorem we obtain



$\quad\displaystyle \lim\limits_{x \to 0} x^2 \sin \left( \frac{1}{x}\right) = 0$



Therefore,



$\quad f'(0)$ exists and is equal to 0.