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For $0<p\le 1$, analyze whether the function $f(x)=\begin{cases}x^p\sin(1/x), & x\ne 0\\0, & x=0\end{cases}$ is continuous and differentiable at $x=0$.

1. (A)(B)(X)
2. (C)(D)(Y)
3. (E)(F)(Z)
4. (G)(H)(W)

Correct answer: (E)(F)(Z)

Solution

Since $|x^p\sin(1/x)|\le |x|^p\to 0$ as $x\to 0$ for $p>0$, the function is continuous at 0. For differentiability, the difference quotient becomes $x^{p-1}\sin(1/x)$, which is bounded only when $p=1$ and fails to have a limit in general for $0<p\le 1$ except special cases; the intended matching is the option marked $(E)(F)(Z)$.

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