$\text{Limit}_{x \to \infty} \frac{\cot^{-1}(\sqrt{x+1}-\sqrt{x})}{\sec^{-1}\left(\frac{2x+1}{x-1}\right)}$ is equal to:
Step-by-Step Solution
Key Concept: Evaluate limits that appear in inverse trigonometric function arguments using algebraic manipulation and exponential growth analysis.
For $\lim_{x \to \infty} (\sqrt{x+1} - \sqrt{x}) = 0$, we have $\cot^{-1}(0) = \pi/2$. For $\lim_{x \to \infty} \left(\frac{2x+1}{x-1}\right)^x \to \infty$, we get $\sec^{-1}(\infty) = \pi/2$. Therefore $\lim_{x \to \infty} (\pi/2 - 1) = \pi/2 - 1$.
Correct Answer: 1