Question

If f(0)=0 and that 'f' is differentiable at x = 0, and ‘k’ is a positive integer. Then ${lim}_{x\to 0}\frac{1}{x}\left[f\right(x)+f\left(\frac{x}{2}\right)+f\left(\frac{x}{3}\right)+\dots \dots +f\left(\frac{x}{k}\right)]$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$l={lim}_{x\to 0}\left\{\begin{array}{c}\frac{f\left(x\right)-f\left(0\right)}{x-0}+\frac{f\left(\frac{x}{2}\right)-f\left(0\right)}{x-0}+\dots \dots +\frac{f\left(\frac{x}{k}\right)-f\left(0\right)}{x-0}\end{array}\right\}(1+\frac{1}{2}+\frac{1}{3}+\dots \dots +\frac{1}{k})f^1\left(0\right)$