Question

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

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

Answer 1

Answer: (B)

$1$

$\underset{x\to \infty }{lim}\frac{co{t}^{-1}\left(x^{-a}{log}_{a}x\right)}{se{c}^{-1}\left(a^x{log}_{x}a\right)}\text{(a > 1 )}$
$=\underset{x\to \infty }{lim}\frac{co{t}^{-1}\left(\frac{{log}_{a}x}{x^a}\right)}{se{c}^{-1}\left(\frac{a^x}{lo{g}_{a}x}\right)}$
Let us find the limit of the term inside the inverse function as $x\to \infty$
$\underset{x\to \infty }{lim}\frac{lo{g}_{a}x}{x^a}$ (Indeterminate form)
Applying L'Hospital's rule,
$=\underset{x\to \infty }{lim}\frac{1}{\left(loga\right)\left(x\right)\left(x^{a-1}\right)}$
$=0$
$\text{As}x\to \infty ,\left(\frac{lo{g}_{a}x}{x^a}\right)\to 0\text{and}\left(\frac{a^x}{lo{g}_{a}x}\right)\to \infty$

Hence, $L=\frac{\pi /2}{\pi /2}=1$