Question

$\underset{x\to \infty }{\text{lim}}\frac{{\cot }^{-1}\left(x^{-a}{\log }_{a}x\right)}{{\sec }^{-1}\left(a^x{\log }_{x}a\right)},(a>1)$ is equal to

• A. $2$
• B. $1$
• C. ${\log }_{a}2$
• D. $0$

Answer 1

The correct option is B $1$

$\underset{x\to \infty }{\text{lim}}\frac{{\cot }^{-1}\left(x^{-a}{\log }_{a}x\right)}{{\sec }^{-1}\left(a^x{\log }_{x}a\right)}=\underset{x\to \infty }{\text{lim}}\frac{{\cot }^{-1}\left(\frac{{\log }_{a}x}{x^a}\right)}{{\sec }^{-1}\left(\frac{a^x}{{\log }_{a}x}\right)}$

Now, $\underset{x\to \infty }{lim}\left(\frac{{\log }_{a}x}{x^a}\right)$
$=\underset{x\to \infty }{lim}\left(\frac{{\log }_{e}x}{x^a}\right)\frac{1}{{\log }_{e}a}$
Using $L$ Hospital rule,
$=\underset{x\to \infty }{lim}\frac{\left({\log }_{a}e\right)\left(\frac{1}{x}\right)}{a{x}^{a-1}}$
$=\underset{x\to \infty }{lim}\frac{{\log }_{a}e}{a{x}^a}=0$

Similarly, $\underset{x\to \infty }{lim}\frac{a^x}{{\log }_{a}x}\to \infty$

$\therefore I=\frac{\frac{\pi }{2}}{\frac{\pi }{2}}=1$