Question

Let $\underset{x\to \infty }{\mathrm{Lim}}{\left({\cot }^{-1}x\right)}^{\frac{1}{x}}=L$; then $L$ is

• A. prime number
• B. composite number
• C. even number divisible by 4
• D. neither prime nor composite

Answer 1

Answer: (D)

$L=e^{{\mathrm{Lim}}_{x\to \infty }\frac{\ln {\cot }^{-1}x}{x}}$ (L'Hospital rule)
$\therefore L=e^{-\infty }=0$