Let `C_(n)= int_(1//n+1)^(1//n)(tan^(-1)(nx))/(sin^(-1) (nx)) dx , "then " lim_(n rarr infty) n^(n). C_(n) ` is equal to

Options

A. 1
B. 0
C. `-1`
D. `(1)/(2)`

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{n \to \infty} n^n C_n \]

where

\[ C_n = \int_{\frac{1}{n+1}}^{\frac{1}{n}} \frac{\tan^{-1}(nx)}{\sin^{-1}(nx)} \, dx. \]

### Step 1: Change of Variables
We start by making a substitution to simplify the integral. Let \( t = nx \). Then, \( dx = \frac{dt}{n} \). The limits change as follows:
- When \( x = \frac{1}{n+1} \), \( t = \frac{n}{n+1} \).
- When \( x = \frac{1}{n} \), \( t = 1 \).

Thus, we can rewrite \( C_n \):

\[ C_n = \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \cdot \frac{dt}{n}. \]

This gives us:

\[ C_n = \frac{1}{n} \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \, dt. \]

### Step 2: Evaluate the Limit
Now we substitute \( C_n \) into the limit:

\[ n^n C_n = n^{n-1} \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \, dt. \]

As \( n \to \infty \), the lower limit \( \frac{n}{n+1} \) approaches \( 1 \). Therefore, we can analyze the behavior of the integral as \( n \to \infty \):

\[ \lim_{n \to \infty} n^{n-1} \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \, dt. \]

### Step 3: Behavior of the Integral
As \( n \to \infty \), the integral can be approximated around \( t = 1 \). We know that:

\[ \tan^{-1}(t) \to \frac{\pi}{4} \quad \text{and} \quad \sin^{-1}(t) \to \frac{\pi}{2} \quad \text{as } t \to 1. \]

Thus, near \( t = 1 \):

\[ \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \to \frac{\frac{\pi}{4}}{\frac{\pi}{2}} = \frac{1}{2}. \]

### Step 4: Final Limit Calculation
The integral becomes:

\[ \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \, dt \approx \frac{1}{2} \cdot \left(1 - \frac{n}{n+1}\right) = \frac{1}{2} \cdot \frac{1}{n+1} \to \frac{1}{2n}. \]

Thus, we have:

\[ n^{n-1} \cdot \frac{1}{2n} = \frac{n^{n-2}}{2}. \]

As \( n \to \infty \), \( n^{n-2} \) diverges to infinity. Therefore, we conclude:

\[ \lim_{n \to \infty} n^n C_n = \frac{1}{2}. \]

### Final Answer
\[ \lim_{n \to \infty} n^n C_n = \frac{1}{2}. \]