Question

$\underset{x\to \infty }{\mathrm{Lim}}\left(x^3\underset{-1/x}{\overset{1/x}{\int }}\frac{\ln \left(1+t^2\right)}{1+e^t}dt\right)$ equals

• A. $1/3$
• B. $2/3$
• C. 1
• D. 0

Answer 1

Answer: (A)

Consider $I=\underset{-1/x}{\overset{1/x}{\int }}\frac{\ln \left(1+t^2\right)}{1+e^t}dt$ ....(1)
$=\underset{-1/x}{\overset{1/x}{\int }}\frac{\ln \left(1+t^2\right)}{1+e^{-t}}dt$ (Using King)
$I=\underset{-1/x}{\overset{1/x}{\int }}\frac{\ln \left(1+t^2\right)e^t}{1+e^t}dt\dots (2)$
(1) + (2)
$2I=\underset{-1/x}{\overset{1/x}{\int }}\ln \left(1+t^2\right)dt=2\underset{0}{\overset{1/x}{\int }}\ln \left(1+t^2\right)dt\Rightarrow I=\underset{0}{\overset{1/x}{\int }}\ln \left(1+t^2\right)dt$
hence $l={\mathrm{Lim}}_{x\to \infty }{x}^3\underset{0}{\overset{1/x}{\int }}\ln \left(1+t^2\right)dt={\mathrm{Lim}}_{x\to \infty }\frac{\underset{0}{\overset{1/x}{\int }}\ln \left(1+t^2\right)dt}{x^{-3}}(\frac{0}{0}$ form $)$
Using L'Hospital's Rule
$l=\underset{x\to \infty }{\mathrm{Lim}}\frac{x^4\ln \left(1+\frac{1}{x^2}\right)\cdot \left(-\frac{1}{x^2}\right)}{-3}=\frac{1}{3}\underset{x\to \infty }{\mathrm{Lim}}{x}^2\ln \left(1+\frac{1}{x^2}\right)=\frac{1}{3}\underset{x\to \infty }{\mathrm{Lim}}\ln {\left(1+\frac{1}{x^2}\right)}^{x^2}(1^{\infty }$ form $)$
$=\underset{x\to \infty }{\mathrm{Lim}}\frac{1}{3}{x}^2\left(1+\frac{1}{x^2}-1\right)=\frac{1}{3}$