Question

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

• A. $2015$
• B. $0$ (zero)
• C. $2015\times 2016$
• D. $2015\times 2014$

Answer 1

The correct option is C $2015\times 2016$

From L Hospital's rule,
${lim}_{x\to 0}\frac{1}{x}[f\left(x\right)+f\left(2x\right)+f\left(3x\right)+...+f\left(2015x\right)]$

$={lim}_{x\to 0}\frac{f^{\prime }\left(x\right)+2f^{\prime }\left(2x\right)+3f^{\prime }\left(3x\right)+...+2015f^{\prime }\left(2015x\right)}{1}$
$=\frac{2\times 2015\times 2016}{2}$
$=2015\times 2016$