Let the functions f, g and h be defined as fo | vedclass

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(C) To check for differentiability at $x = 0$,we calculate the derivative using the limit definition $f'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h}$.

Vedclass · Accepted Answer

$g$ and $h$ only

Vedclass · Answer

(C) To check for differentiability at $x = 0$,we calculate the derivative using the limit definition $f'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h}$.$(1)$ For $f(x)$:$f'(0) = \lim_{h 	o 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h 	o 0} \sin(1/h)$.Since $\lim_{h 	o 0} \sin(1/h)$ oscillates between $-1$ and $1$,the limit does not exist. Thus,$f(x)$ is not differentiable at $x = 0$.$(2)$ For $g(x)$:$g'(0) = \lim_{h 	o 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h 	o 0} h \sin(1/h)$.Since $|h \sin(1/h)| \le |h|$,by the Squeeze Theorem,the limit is $0$. Thus,$g(x)$ is differentiable at $x = 0$.$(3)$ For $h(x) = |x|^3$:$h'(0) = \lim_{h 	o 0} \frac{|h|^3 - 0}{h} = \lim_{h 	o 0} \frac{|h|^3}{h}$.For $h > 0$,$\frac{h^3}{h} = h^2 	o 0$. For $h Since the limit exists and is $0$,$h(x)$ is differentiable at $x = 0$.Therefore,$g$ and $h$ are differentiable at $x = 0$. The correct option is $C$.

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