Question

The value of $\underset{x\to \infty }{lim}{x}^2\ln \left(x{\cot }^{-1}x\right)$ is

• A. $-\frac{1}{3}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $-\frac{2}{3}$

Answer 1

The correct option is A $-\frac{1}{3}$

$\underset{x\to \infty }{lim}{x}^2\ln \left(x{\cot }^{-1}x\right)$
Assuming $y=\frac{1}{x}$, we get
$\underset{x\to \infty }{lim}{x}^2\ln \left(x{\cot }^{-1}x\right)=\underset{y\to 0}{lim}\frac{1}{y^2}\ln \left(\frac{1}{y}{\cot }^{-1}\frac{1}{y}\right)=\underset{y\to 0}{lim}\frac{\ln \left(\frac{1}{y}{\tan }^{-1}y\right)}{y^2}=\underset{y\to 0}{lim}\frac{\ln \left(\frac{{\tan }^{-1}y}{y}\right)}{y^2}=\underset{y\to 0}{lim}\frac{\ln \left({\tan }^{-1}y\right)-\ln y}{y^2}$
Using L'Hospital's Rule, we get
$=\underset{y\to 0}{lim}\frac{\frac{1}{(1+y^2)\left({\tan }^{-1}y\right)}-\frac{1}{y}}{2y}=\underset{y\to 0}{lim}\frac{y-(1+y^2)\left({\tan }^{-1}y\right)}{2y^3(1+y^2)\frac{\left({\tan }^{-1}y\right)}{y}}=\underset{y\to 0}{lim}\frac{y-(1+y^2)\left({\tan }^{-1}y\right)}{2y^3}$
Using L'Hospital's Rule, we get
$=\underset{y\to 0}{lim}\frac{1-1-2y{\tan }^{-1}y}{6y^2}=\underset{y\to 0}{lim}\frac{-{\tan }^{-1}y}{3y}=-\frac{1}{3}$