Prove that the function f(x) = x sin(1/x) for x ≠ 0 and f(0) = 0 is continuous everywhere but not differentiable at $x = 0$.

Step 1: Check continuity for $x \neq 0$

For $x \neq 0$, the function $f(x) = x \sin(1/x)$ is a product of two continuous functions: $g(x) = x$ and $h(x) = \sin(1/x)$. The function $g(x)=x$ is continuous everywhere. The function $h(x) = \sin(1/x)$ is a composition of $\sin(u)$ and $u=1/x$. Since $1/x$ is continuous for $x \neq 0$ and $\sin(u)$ is continuous everywhere, their composition is continuous for $x \neq 0$. Therefore, their product $f(x)$ is continuous for all $x \neq 0$.

Step 2: Check continuity at $x = 0$

To check continuity at $x = 0$, we need to evaluate the limit of $f(x)$ as $x$ approaches $0$. We know that the sine function is bounded between $-1$ and $1$, i.e., $-1 \le \sin(1/x) \le 1$. Multiplying this inequality by $|x|$ (which is positive), we get $-|x| \le x \sin(1/x) \le |x|$. As $x$ approaches $0$, $|x|$ approaches $0$. By the Squeeze Theorem, $\lim_{x \to 0} x \sin(1/x) = 0$. Since $f(0)$ is given as $0$, we have $\lim_{x \to 0} f(x) = f(0)$, which means the function is continuous at $x = 0$.

Step 3: Check differentiability at $x = 0$

To check differentiability at $x = 0$, we use the definition of the derivative: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$. Substituting the function definition, we get $\lim_{h \to 0} \frac{h \sin(1/h) - 0}{h}$. This simplifies to $\lim_{h \to 0} \sin(1/h)$.

Step 4: Evaluate the limit for differentiability

As $h$ approaches $0$, $1/h$ approaches $\pm \infty$. The function $\sin(u)$ oscillates between $-1$ and $1$ as $u$ approaches $\pm \infty$. Therefore, $\lim_{h \to 0} \sin(1/h)$ does not approach a single value and thus does not exist. Since the limit does not exist, the function is not differentiable at $x = 0$.

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