`lim_(n to oo)n^2*int_(1/(n+1))^(1/n)(tan^(-1)(nx)/sin^(-1)(nx)dx)` is equal to

Answer

To solve the limit

\[ \lim_{n \to \infty} n^2 \int_{\frac{1}{n+1}}^{\frac{1}{n}} \frac{\tan^{-1}(nx)}{\sin^{-1}(nx)} \, dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( t = nx \). Then, we have:
\[ dx = \frac{dt}{n}. \]
When \( x = \frac{1}{n+1} \), \( t = \frac{n}{n+1} \) and when \( x = \frac{1}{n} \), \( t = 1 \).

### Step 2: Change of Limits and Rewrite the Integral
Substituting into the integral, we get:
\[ \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \cdot \frac{dt}{n}. \]
Thus, the limit becomes:
\[ \lim_{n \to \infty} n^2 \cdot \frac{1}{n} \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \, dt = \lim_{n \to \infty} n \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \, dt. \]

### Step 3: Evaluate the Integral
As \( n \to \infty \), the lower limit \( \frac{n}{n+1} \to 1 \). Therefore, we can write:
\[ \lim_{n \to \infty} n \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \, dt. \]
This integral approaches:
\[ \int_{1}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \, dt = 0. \]

### Step 4: Analyze the Behavior Near the Limit
To analyze the behavior near the limit, we can apply L'Hôpital's Rule since we have a \( \frac{0}{0} \) form. We differentiate the numerator and denominator separately.

### Step 5: Applying L'Hôpital's Rule
Differentiate the integral with respect to \( n \):
\[ \text{Numerator: } \frac{d}{dn} \left( n \int_{\frac{n}{n+1}}^{1} \frac{\tan^{-1}(t)}{\sin^{-1}(t)} \, dt \right), \]
and the denominator:
\[ \text{Denominator: } \frac{d}{dn} \left( n + 1 \right). \]

### Step 6: Final Calculation
After applying L'Hôpital's Rule and simplifying, we find:
\[ \lim_{n \to \infty} \frac{n \cdot \frac{\tan^{-1}(1)}{\sin^{-1}(1)}}{1} = \frac{\frac{\pi}{4}}{\frac{\pi}{2}} = \frac{1}{2}. \]

Thus, the final answer is:
\[ \boxed{\frac{1}{2}}. \]