Question

$f\left(x\right)=\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{n}{n^2+k^2{x}^2}$, $x>0$ is equal to

• A. $x{\tan }^{-1}\left(x\right)$
• B. ${\tan }^{-1}\left(x\right)$
• C. $\frac{{\tan }^{-1}\left(x\right)}{x}$
• D. none of the above

Answer 1

The correct option is C $\frac{{\tan }^{-1}\left(x\right)}{x}$

$f\left(x\right)=\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{n}{n^2+k^2{x}^2},x>0$

Let $\frac{k}{n}=y,dy=\frac{1}{n}$

$=\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{dy}{1+x^2{y}^2}$

limit of $y$ is $0$ to $1$

$={\int }_0^1\frac{1}{1+y^2{x}^2}dy$

Let $xy=t$

$\Rightarrow xdy=dt$

$=\frac{1}{x}{\int }_0^1\frac{1}{1+t^2}dt$

$=\frac{1}{x}[{\tan }^{-1}t{]}_0^1=\frac{1}{x}[{\tan }^{-1}xy{]}_0^1$

$=\frac{1}{x}[{\tan }^{-1}x-{\tan }^{-1}(0\left)\right]$

$=\frac{{\tan }^{-1}x}{x}$

$\therefore f\left(x\right)=\frac{{\tan }^{-1}x}{x}$