Question

Suppose the limit
$L=\underset{n\to \infty }{lim}\sqrt{n}\underset{0}{\overset{1}{\int }}\frac{1}{(1+x^2{)}^n}dx$
exists and is larger than $\frac{1}{2}$. Then

• A. $\frac{1}{2}<L<2$
• B. $2<L<3$
• C. $3<L<4$
• D. $L\ge 4$

Answer 1

The correct option is A $\frac{1}{2}<L<2$

Assuming $I=\underset{0}{\overset{1}{\int }}\frac{1}{(1+x^2{)}^n}dx$
From binomial expansion we can say,
$(1+x^2{)}^n>(1+n{x}^2)\Rightarrow \frac{1}{(1+x^2{)}^n}<\frac{1}{1+n{x}^2}$
so,
$\Rightarrow \underset{0}{\overset{1}{\int }}\frac{1}{(1+x^2{)}^n}dx<\underset{0}{\overset{1}{\int }}\frac{1}{1+n{x}^2}dx$
$\Rightarrow I<\frac{1}{\sqrt{n}}{\left[{\tan }^{-1}\sqrt{n}x\right]}_0^1$
$\Rightarrow I<\frac{1}{\sqrt{n}}\left[{\tan }^{-1}\sqrt{n}\right]$

$\therefore L=\underset{n\to \infty }{lim}\sqrt{n}I<\underset{n\to \infty }{lim}\sqrt{n}\frac{1}{\sqrt{n}}\left[{\tan }^{-1}\sqrt{n}\right]$
$\Rightarrow L<\frac{\pi }{2}$

Hence $\frac{1}{2}<L<2$