Given f(x) = cases x²sin 1/x, x ≠ 0,, [6pt] 0, x = 0

Given $f(x) = \begin{cases} x^{2}\sin\dfrac{1}{x}, & x \neq 0, \\[6pt] 0, & x = 0, \end{cases}$ find $f'_{-}(0)$ and $f'_{+}(0)$. Does $f'(0)$ exist?

Step 1: Form the difference quotient at $x = 0$

For $\Delta x \neq 0$, using $f(0) = 0$: \[\frac{f(0+\Delta x)-f(0)}{\Delta x} = \frac{(\Delta x)^{2}\sin\dfrac{1}{\Delta x}}{\Delta x} = \Delta x\,\sin\frac{1}{\Delta x}.\]

One factor $\Delta x$ cancels, but a second one survives — this is the decisive difference from Problem 21.

Step 2: Evaluate the limit by the squeeze theorem

Since $\left|\sin\dfrac{1}{\Delta x}\right| \leq 1$ for every $\Delta x \neq 0$: \[0 \leq \left|\Delta x\,\sin\frac{1}{\Delta x}\right| \leq |\Delta x| \;\xrightarrow[\Delta x\to 0]{}\; 0.\]

By the squeeze theorem, $\lim_{\Delta x \to 0} \Delta x\,\sin\frac{1}{\Delta x} = 0$.

Step 3: Compute the one-sided derivatives

The estimate above uses only $|\Delta x|$, so it is valid from either side: \[f'_{-}(0) = \lim_{\Delta x \to 0^{-}} \Delta x\,\sin\frac{1}{\Delta x} = 0, \qquad f'_{+}(0) = \lim_{\Delta x \to 0^{+}} \Delta x\,\sin\frac{1}{\Delta x} = 0.\]

Step 4: Conclude on the existence of $f'(0)$

Both one-sided derivatives exist and are equal, so the two-sided derivative exists and takes the same value: $f'(0) = \lim_{\Delta x \to 0} \Delta x\,\sin\frac{1}{\Delta x} = 0$.

Step 5: Verification

Away from the origin the product and chain rules give

\[f'(x) = 2x\sin\frac{1}{x} - \cos\frac{1}{x}, \qquad x \neq 0\]

\[\Delta x = 10^{-3}:\quad \Delta x\,\sin\frac{1}{\Delta x} = 10^{-3}\sin(1000) \approx 8.27\times 10^{-4} \to 0 \;\;\checkmark\]