A function f(x) is defined as f(x) = begin{ca | vedclass

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(C) First,we find $f'(0)$ using the definition of the derivative: $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^m \sin(1/h) -

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$3$

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(C) First,we find $f'(0)$ using the definition of the derivative: $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^m \sin(1/h) - 0}{h} = \lim_{h 	o 0} h^{m-1} \sin(1/h)$. For this limit to exist and be equal to $0$,we must have $m-1 > 0$,i.e.,$m > 1$. Now,for $x 
eq 0$,$f'(x) = m x^{m-1} \sin(1/x) - x^{m-2} \cos(1/x)$. For $f'(x)$ to be continuous at $x = 0$,we require $\lim_{x 	o 0} f'(x) = f'(0) = 0$. $\lim_{x 	o 0} [m x^{m-1} \sin(1/x) - x^{m-2} \cos(1/x)] = 0$. The first term $m x^{m-1} \sin(1/x)$ approaches $0$ if $m-1 > 0$,i.e.,$m > 1$. The second term $x^{m-2} \cos(1/x)$ approaches $0$ if $m-2 > 0$,i.e.,$m > 2$. Since $m \in N$,the smallest integer $m$ satisfying $m > 2$ is $m = 3$.

$A$ function $f(x)$ is defined as $f(x) = \begin{cases} x^m \sin \frac{1}{x} & x \neq 0, m \in N \\ 0 & x = 0 \end{cases}$. The least value of $m$ for which $f'(x)$ is continuous at $x = 0$ is

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