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(C) First,check differentiability at $x=0$: $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{2-2h^2-h^2 \sin(1/h) - 2}{h} = \lim_{

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For any positive real number $\delta$,the function $f$ is not an increasing function on the interval $(-\delta, 0)$

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(C) First,check differentiability at $x=0$: $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{2-2h^2-h^2 \sin(1/h) - 2}{h} = \lim_{h 	o 0} (-2h - h \sin(1/h)) = 0$. Since the limit exists,$f$ is differentiable at $x=0$. Thus,$(A)$ is false. For $x 
eq 0$,$f'(x) = -4x - 2x \sin(1/h) - x^2 \cos(1/h) (-1/x^2) = -4x - 2x \sin(1/x) + \cos(1/x)$. As $x 	o 0$,$f'(x)$ oscillates due to the $\cos(1/x)$ term. For any $\delta > 0$,in the interval $(0, \delta)$ or $(-\delta, 0)$,$f'(x)$ takes both positive and negative values because $\cos(1/x)$ oscillates between $-1$ and $1$. Therefore,$f$ is neither increasing nor decreasing on any interval $(0, \delta)$ or $(-\delta, 0)$. This makes $(B)$ false and $(C)$ true. Finally,$f(0)=2$ and for small $h 
eq 0$,$f(h) = 2 - 2h^2 - h^2 \sin(1/h) = 2 - h^2(2 + \sin(1/h))$. Since $2 + \sin(1/h) > 0$,$f(h) < 2$ for small $h$. Thus,$x=0$ is a local maximum,making $(D)$ false.

Let $R$ denote the set of all real numbers. Define the function $f: R \rightarrow R$ by $f(x) = \begin{cases} 2-2x^2-x^2 \sin \frac{1}{x} & \text{if } x \neq 0 \\ 2 & \text{if } x=0 \end{cases}$. Then which one of the following statements is True?

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