`lim_(xrarroo) (cot^(-1)(sqrt(x+1)+sqrtx))/(sec^(-1){((2x+1)/(x-1))^(x)})=`

Options

A. 1
B. `0`
C. `pi//2`
D. non existent

Explanation

To solve the limit

\[ \lim_{x \to \infty} \frac{\cot^{-1}(\sqrt{x+1} + \sqrt{x})}{\sec^{-1}\left(\left(\frac{2x+1}{x-1}\right)^x\right)}, \]

we will analyze the numerator and denominator separately.

### Step 1: Analyze the Numerator

The numerator is \(\cot^{-1}(\sqrt{x+1} + \sqrt{x})\).

1. As \(x \to \infty\), we can simplify \(\sqrt{x+1}\):
\[ \sqrt{x+1} = \sqrt{x(1 + \frac{1}{x})} = \sqrt{x} \sqrt{1 + \frac{1}{x}} \approx \sqrt{x} \left(1 + \frac{1}{2x}\right) \text{ (using Taylor expansion)} \]
Thus,
\[ \sqrt{x+1} \approx \sqrt{x} + \frac{1}{2\sqrt{x}}. \]
Therefore,
\[ \sqrt{x+1} + \sqrt{x} \approx 2\sqrt{x} + \frac{1}{2\sqrt{x}}. \]

2. As \(x \to \infty\), \(\sqrt{x+1} + \sqrt{x} \to 2\sqrt{x}\).

3. Now, we take the limit:
\[ \cot^{-1}(2\sqrt{x}) \to \cot^{-1}(\infty) = 0. \]

### Step 2: Analyze the Denominator

The denominator is \(\sec^{-1}\left(\left(\frac{2x+1}{x-1}\right)^x\right)\).

1. Simplifying the expression inside the secant:
\[ \frac{2x+1}{x-1} = \frac{2 + \frac{1}{x}}{1 - \frac{1}{x}} \to 2 \text{ as } x \to \infty. \]
Therefore,
\[ \left(\frac{2x+1}{x-1}\right)^x \to 2^x. \]

2. Now, we compute the limit:
\[ \sec^{-1}(2^x) \to \sec^{-1}(\infty) = \frac{\pi}{2}. \]

### Step 3: Combine the Results

Now we can combine the results from the numerator and denominator:

\[ \lim_{x \to \infty} \frac{\cot^{-1}(\sqrt{x+1} + \sqrt{x})}{\sec^{-1}\left(\left(\frac{2x+1}{x-1}\right)^x\right)} = \frac{0}{\frac{\pi}{2}} = 0. \]

### Final Answer

Thus, the limit is

\[ \boxed{0}. \]