91Ó°ÊÓ

We first simplify the expressions inside the inverse trigonometric functions: 1. \(\cot ^{-1}\left(\sqrt{x-1} - \sqrt{x}\right)\) 2. \(\sec ^{-1}\left(\left(\frac{2x+1}{x-1}\right)^x\right)\) Note that we don't have to simplify the first one, but we will simplify the second expression by taking natural logarithm on both sides. Let \(y = \sec ^{-1}\left(\left(\frac{2x+1}{x-1}\right)^x\right)\), then \(\sec(y) = \left(\frac{2x+1}{x-1}\right)^x\) Taking natural logarithm on both sides, \(\ln(\sec(y)) = x \ln\left(\frac{2x+1}{x-1}\right)\)

Replace cotangent inverse with the tangent inverse, using the property: \(\cot^{-1}(x) = \tan^{-1}(\frac{1}{x})\) Therefore, the expression becomes: \(\lim_{x\to\infty}\frac{\tan^{-1}\left(\frac{1}{\sqrt{x-1} - \sqrt{x}}\right)}{\sec^{-1}\left(\left(\frac{2x+1}{x-1}\right)^x\right)}\)

We have the form \(\frac{0}{0}\) as \(x \rightarrow \infty\), which means we can apply L'Hôpital's Rule. First, find the derivatives of the numerator and denominator with respect to x: \(\frac{d}{dx}\tan^{-1}\left(\frac{1}{\sqrt{x-1} - \sqrt{x}}\right) = \frac{\frac{1}{2}\left(\frac{1}{x} - \frac{1}{x-1}\right)}{\left(\frac{1}{\sqrt{x-1} - \sqrt{x}}\right)^2 + 1}\) \(\frac{d}{dx}\ln(\sec(y)) = \frac{2}{2x+1} - \frac{1}{x-1}\) Now, get the limit of the derivative of the numerator divided by the derivative of the denominator as \(x \rightarrow \infty\): \(\lim _{x \rightarrow \infty} \frac{\frac{1}{2}\left(\frac{1}{x} - \frac{1}{x-1}\right)}{\left(\frac{1}{\sqrt{x-1} - \sqrt{x}}\right)^2 + 1} \div \lim_{x \rightarrow \infty} \frac{\frac{2}{2x+1} - \frac{1}{x-1}}{1} = 0\) Since the limit equals 0, the correct answer is (B) 0.