Question

Let $f:[0,1]\to R$ be a differentiable function with $f(0)=0$, then $\underset{n\to \infty }{\mathrm{Lim}}{n}^2\underset{0}{\overset{\frac{1}{n}}{\int }}f(t)dt$ equals

• A. $\frac{1}{2}{f}^{\prime }(0)$
• B. $f^{\prime }(0)$
• C. $2f^{\prime }(0)$
• D. 0

Answer 1

Answer: (A)

Let $I=\underset{n\to \infty }{\mathrm{Lim}}{n}^2\underset{0}{\overset{\frac{1}{n}}{\int }}f(t)dt=\underset{n\to \infty }{\mathrm{Lim}}\frac{\underset{0}{\overset{1/n}{\int }}f(t)dt}{\frac{1}{n^2}}(\frac{0}{0}$ form, as $n\to \infty )$
Using L'Hospital's rule and Using Leibnitz rule, we get
$I=\underset{n\to \infty }{\mathrm{Lim}}\frac{n^{3}f\left(\frac{1}{n}\right)\left(\frac{-1}{n^2}\right)}{-2}=\underset{n\to \infty }{\mathrm{Lim}}\frac{n}{2}f\left(\frac{1}{n}\right)$
Put $n=\frac{1}{h}$, hence
$I=\frac{1}{2}\underset{n\to \infty }{\mathrm{Lim}}\left(\frac{f(0+h)-f(0)}{h}\right)=\frac{1}{2}{f}^{\prime }(0)$