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(C) Given,$f(0)=0$ and $f'(0)=2$. Using $L$'Hopital's rule for the limit $\lim _{x \rightarrow 0} \frac{\sum_{k=1}^{2015} f(kx)}{x}$: $= \lim _{x \rig

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$2015 	imes 2016$

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(C) Given,$f(0)=0$ and $f'(0)=2$. Using $L$'Hopital's rule for the limit $\lim _{x ightarrow 0} \frac{\sum_{k=1}^{2015} f(kx)}{x}$: $= \lim _{x ightarrow 0} \frac{\frac{d}{dx} [f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)]}{1}$ $= \lim _{x ightarrow 0} [f'(x) + 2f'(2x) + 3f'(3x) + \ldots + 2015f'(2015x)]$ $= f'(0) + 2f'(0) + 3f'(0) + \ldots + 2015f'(0)$ $= f'(0) [1 + 2 + 3 + \ldots + 2015]$ $= 2 	imes \frac{2015(2015+1)}{2}$ $= 2015 	imes 2016$.

Let $f: R \rightarrow R$ be differentiable at $x=0$. If $f(0)=0$ and $f'(0)=2$,then the value of $\lim _{x \rightarrow 0} \frac{1}{x} [f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)]$ is

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