Question

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

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

Answer 1

The correct option is C $2015\times 2016$

Given, $f\left(0\right)=0$ and $f^{\prime }\left(0\right)=2$

Therefore, $\underset{x\to 0}{lim}\frac{1}{x}\left[f\right(x)+f(2x)+f(3x)+...+f(2015x\left)\right]$

$=\underset{x\to 0}{lim}\frac{\left[f\right(x)+2f(2x)+3f(3x)+...+2015f(2015x\left)\right]}{1}$ ....[Applying L Hospitals rule]

$=\frac{2+2\times 2+3\times 2+...+2015\times 2016}{1}$

$=2[1+2+3+...+2015]$

$=\frac{2\left(2015\right)(2015+1)}{2}=\frac{2\times 2015\times 2016}{2}$

$=2015\times 2016$