The value of $\displaystyle\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$ for $(a>1)$ is equal to?

$\mathrm{L} \displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$

$\displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{\log{x}}{\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log a}{\log x}}}}=\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{1}{a\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log^2 a. x}{1}}}}=\frac{\cot^{-1}0}{\sec^{-1} \infty}=\frac{\pi/2}{\pi/2}=1$

Note: L-Hospital's rule has been used in step two in both numerator and denominator alone.