Question

If f : R $\to$ R be a differentiable function and f(0) = 0 and f'(0) = 1 then ${lim}_{x\to 0}\frac{1}{x}\left[f\right(x)+f(\frac{x}{2})+...+f(\frac{x}{100}\left)\right]$ equals

• A. $1$
• B. $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}$
• C. ${}^{100}{C}_1-\frac{{}^{100}{C}_2}{2}+\frac{{}^{100}{C}_3}{3}+...+(-1{)}^{99}\frac{{}^{100}{C}_{100}}{100}$
• D. Does not exist

Answer 1

The correct option is A $1$

$f\left(x\right)=0$ $f^{\prime }\left(x\right)=1$

Let $L={lim}_{x\to 0}\frac{f\left(x\right)+f\left(\frac{x}{2}\right)+........+f\left(\frac{x}{100}\right)}{x}$

Applying L'hospital's rule

$\Rightarrow L={lim}_{x\to 0}\frac{f^{\prime }\left(x\right)+\frac{1}{2}{f}^{\prime }\left(\frac{x}{2}\right)+........+\frac{1}{100}{f}^{\prime }\left(\frac{x}{100}\right)}{1}$

$=f^{\prime }\left(0\right)+\frac{1}{2}{f}^{\prime }\left(\frac{0}{2}\right)+......\frac{1}{100}{f}^{\prime }\left(\frac{0}{100}\right)$

$=1+\frac{1}{2}+\frac{1}{3}+.....\frac{1}{100}$

( Since $f^{\prime }\left(0\right)=1$ )

Hence, the answer is $1.$