Let f(x) = begin{cases} x^p sin frac{1}{x}, & | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 0$.Since $\lim_{x 	o 0} x^p \sin \frac{1}{x} = 0$ only if $p >

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$0 < p \le 1$

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 0$.Since $\lim_{x 	o 0} x^p \sin \frac{1}{x} = 0$ only if $p > 0$,the function is continuous for $p > 0$.For $f(x)$ to be differentiable at $x = 0$,the limit $f'(0) = \lim_{x 	o 0} \frac{f(x) - f(0)}{x - 0}$ must exist.$f'(0) = \lim_{x 	o 0} \frac{x^p \sin \frac{1}{x} - 0}{x} = \lim_{x 	o 0} x^{p-1} \sin \frac{1}{x}$.This limit exists and equals $0$ if $p - 1 > 0$,i.e.,$p > 1$.If $p \le 1$,the limit $\lim_{x 	o 0} x^{p-1} \sin \frac{1}{x}$ does not exist.Therefore,$f(x)$ is continuous but not differentiable at $x = 0$ when $0 < p \le 1$.

Let $f(x) = \begin{cases} x^p \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Then $f(x)$ is continuous but not differentiable at $x = 0$ if:

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