Question

If $L=\underset{n\to \infty }{lim}\underset{a}{\overset{\infty }{\int }}\frac{ndx}{1+n^2{x}^2}$, where $a\in R$, then the value of $L$ can be

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $0$
• D. $1$

Answer 1

Answer: (A), (B), (C)

$0$

Consider,
$I=\underset{a}{\overset{\infty }{\int }}\frac{ndx}{n^2(x^2+\frac{1}{n^2})}=\frac{1}{n}\cdot n{\left({\tan }^{-1}nx\right)}_a^{\infty }=(\frac{\pi }{2}-{\tan }^{-1}an)\therefore L=\underset{n\to \infty }{lim}(\frac{\pi }{2}-{\tan }^{-1}an)=[\begin{array}{cc}\pi ,& \text{if}a<0\\ \frac{\pi }{2},& \text{if}a=0\\ 0,& \text{if}a>0\end{array}$