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(C) To check the differentiability of $f(x)$ at $x=0$,we examine the limit of the difference quotient:$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} =

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$\{x \in [-1,1]: x=\frac{2}{2n+1}, n \in Z\}$

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(C) To check the differentiability of $f(x)$ at $x=0$,we examine the limit of the difference quotient:$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^2 \left| \cos \left(\frac{\pi}{h}ight) ight| - 0}{h} = \lim_{h 	o 0} h \left| \cos \left(\frac{\pi}{h}ight) ight|$.Since $|\cos(\frac{\pi}{h})| \leq 1$,by the Squeeze Theorem,$\lim_{h 	o 0} h \left| \cos \left(\frac{\pi}{h}ight) ight| = 0$. Thus,$f$ is differentiable at $x=0$.For $x 
eq 0$,$f(x) = x^2 |\cos(\frac{\pi}{x})|$. The function $|\cos(\frac{\pi}{x})|$ is not differentiable where the argument of the absolute value is zero,i.e.,$\cos(\frac{\pi}{x}) = 0$.$\cos(\frac{\pi}{x}) = 0 \implies \frac{\pi}{x} = (2n+1)\frac{\pi}{2}$ for $n \in Z$.$x = \frac{2}{2n+1}$.Thus,$f$ is not differentiable at points where $\cos(\frac{\pi}{x}) = 0$,which are $x = \frac{2}{2n+1}$ for $n \in Z$.

Let $f:[-1,1] \rightarrow R$ be a function defined by $f(x)=\begin{cases} x^2 \left| \cos \left(\frac{\pi}{x}\right) \right| & \text{for } x \neq 0 \\ 0 & \text{for } x=0 \end{cases}$. The set of points where $f$ is not differentiable is

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