Question

$\underset {x \rightarrow \infty}{\text{Lim}} \left(x^3 \int\limits_{-1 / x}^{1 / x} \frac{\ln \left(1+t^2\right)}{1+e^t} d t\right)$ equals

• A. $1 / 3$
• B. $2 / 3$
• C. 1
• D. 0

Answer 1

Answer: (A)

Consider $\quad I =\int\limits_{-1 / x }^{1 / x } \frac{\ln \left(1+ t ^2\right)}{1+ e ^{ t }} dt$ ....(1)
$=\int\limits_{-1 / x}^{1 / x} \frac{\ln \left(1+ t ^2\right)}{1+ e ^{- t }} dt \quad$ (Using King)
$I =\int\limits_{-1 / x }^{1 / x } \frac{\ln \left(1+ t ^2\right) e ^{ t }}{1+ e ^{ t }} dt \ldots(2)$
(1) + (2)
$2 I =\int\limits_{-1 / x }^{1 / x } \ln \left(1+ t ^2\right) dt =2 \int\limits_0^{1 / x } \ln \left(1+ t ^2\right) dt \quad \Rightarrow \quad I =\int\limits_0^{1 / x } \ln \left(1+ t ^2\right) dt$
hence $l=\operatorname{Lim}_{x \rightarrow \infty} x^3 \int\limits_0^{1 / x} \ln \left(1+t^2\right) dt =\operatorname{Lim}_{x \rightarrow \infty} \frac{\int\limits_0^{1 / x} \ln \left(1+t^2\right) dt }{x^{-3}} \quad\left(\frac{0}{0}\right.$ form $)$
Using L'Hospital's Rule
$l=\underset{x \rightarrow \infty} {\text{Lim}} \frac{x^4 \ln \left(1+\frac{1}{x^2}\right) \cdot\left(-\frac{1}{x^2}\right)}{-3}=\frac{1}{3} \underset{x \rightarrow \infty} {\text{Lim}}x^2 \ln \left(1+\frac{1}{x^2}\right)=\frac{1}{3} \underset{x \rightarrow \infty} {\text{Lim}} \ln \left(1+\frac{1}{x^2}\right)^{x^2}\left(1^{\infty}\right.$ form $)$
$=\underset{x \rightarrow \infty} {\text{Lim}} \frac{1}{3} x^2\left(1+\frac{1}{x^2}-1\right)=\frac{1}{3}$