Question

$\underset{x\to \infty }{lim}{\int }_0^{x^{}/2}\frac{t^2}{x^2(1+t^2)}dt$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

$\underset{x\to \infty }{lim}{\int }_0^{x/2}\frac{t^2}{x^2(1+t^2)}dt=\underset{x\to \infty }{lim}\frac{1}{x^2}{\int }_0^{x/2}\frac{t^2}{1+t^2}dt$

$=\underset{x\to \infty }{lim}\frac{1}{x^2}\left[{\int }_0^{x/2}\frac{t^2+1-1}{1+t^2}dt\right]$

$=\underset{x\to \infty }{lim}\frac{1}{x^2}\left[{\int }_0^{x/2}\frac{1+t^2}{1+t^2}dt-{\int }_0^{x/2}\frac{1}{1+t^2}dt\right]$

$=\underset{x\to \infty }{lim}\frac{1}{x^2}\left[\right[t{]}_0^{x/2}-\left[{\tan }^{-1}t{]}_0^{x/2}\right]$

$=\underset{x\to \infty }{lim}\frac{1}{x^2}[\frac{x}{2}-{\tan }^{-1}\frac{x}{2}+{\tan }^{-1}0]$

$=\underset{x\to \infty }{lim}\frac{1}{x^2}[\frac{x}{2}-{\tan }^{-1}\frac{x}{2}]$

$=\underset{x\to \infty }{lim}[\frac{1}{2x}-\frac{1}{x^2}{\tan }^{-1}\frac{x}{2}]$

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

$=0-\frac{1}{\infty }\cdot \frac{\pi }{2}$

$=0-0$

$=0$.