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

Given $f(x) = \begin{cases} x\sin\dfrac{1}{x}, & x \neq 0, \\[6pt] 0, & x = 0, \end{cases}$ does $f'(0)$ exist? Discuss the continuity of $f$.

Step 1: Check continuity at $x = 0$

Since $\left|\sin\dfrac{1}{x}\right| \leq 1$ for every $x \neq 0$, the values of $f$ are squeezed against zero: \[\bigl|f(x) - f(0)\bigr| = \left|x\sin\frac{1}{x}\right| = |x|\left|\sin\frac{1}{x}\right| \leq |x| \;\xrightarrow[x\to 0]{}\; 0.\]

Hence $\displaystyle\lim_{x\to 0} f(x) = 0 = f(0)$, so $f$ is continuous at $x = 0$. Away from the origin $f$ is a product and composition of continuous functions, so it is continuous on all of $\mathbb{R}$.

Step 2: Form the difference quotient at $x = 0$

\[ \frac{f(0+\Delta x)-f(0)}{\Delta x} = \frac{\Delta x\,\sin\dfrac{1}{\Delta x}}{\Delta x} = \sin\frac{1}{\Delta x}. \]

The factor $\Delta x$ cancels completely, leaving a purely oscillating expression.

Step 3: Investigate the limit of the difference quotient

As $\Delta x \to 0$ the quantity $\sin\dfrac{1}{\Delta x}$ sweeps through the whole interval $[-1,1]$ infinitely often. To prove that no limit exists, choose two sequences tending to zero:

\[\Delta x_{k} = \frac{1}{2\pi k} \;\Longrightarrow\; \sin\frac{1}{\Delta x_{k}} = \sin(2\pi k) = 0\]

\[\Delta x_{k} = \frac{1}{\frac{\pi}{2}+2\pi k} \;\Longrightarrow\; \sin\frac{1}{\Delta x_{k}} = \sin\!\left(\frac{\pi}{2}+2\pi k\right) = 1\]

The difference quotient tends to $0$ along the first sequence and to $1$ along the second. Since the value of the limit would have to be the same along every sequence, no limit exists.

Step 4: Conclude on differentiability

\[ f'(0) = \lim_{\Delta x \to 0}\sin\frac{1}{\Delta x} \quad\text{does not exist,} \]

and neither one-sided derivative exists either, since the same two sequences can be chosen on each side of the origin.