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

Answer

To solve the limit

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

we will follow these steps:

### Step 1: Change of Variables
Let \( nx = t \). Then, \( dx = \frac{dt}{n} \). The limits of integration 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 the integrals:

\[ \int_{\frac{1}{n+1}}^{\frac{1}{n}} \tan^{-1}(nx) \, dx = \int_{\frac{n}{n+1}}^{1} \tan^{-1}(t) \cdot \frac{dt}{n} \]

and

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

### Step 2: Substitute into the Limit
Now substituting these into our limit expression:

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

### Step 3: Evaluate the Limit
As \( n \to \infty \), the lower limit \( \frac{n}{n+1} \) approaches \( 1 \). Therefore, we can evaluate the integrals:

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

Both integrals approach \( 0 \) as the lower limit approaches \( 1 \).

### Step 4: Apply L'Hôpital's Rule
Since both the numerator and denominator approach \( 0 \), we can apply L'Hôpital's Rule:

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

Using Leibniz's rule, we differentiate under the integral sign.

### Step 5: Differentiate the Integrals
The derivative of the numerator becomes:

\[ -\tan^{-1}\left(\frac{n}{n+1}\right) \cdot \frac{1}{(n+1)^2} \]

and for the denominator:

\[ -\sin^{-1}\left(\frac{n}{n+1}\right) \cdot \frac{1}{(n+1)^2}. \]

### Step 6: Evaluate the Final Limit
Thus, we have:

\[ \lim_{n \to \infty} \frac{\tan^{-1}\left(\frac{n}{n+1}\right)}{\sin^{-1}\left(\frac{n}{n+1}\right)}. \]

As \( n \to \infty \), both \( \tan^{-1}(1) \) and \( \sin^{-1}(1) \) approach \( \frac{\pi}{4} \) and \( \frac{\pi}{2} \) respectively.

Thus, the limit simplifies to:

\[ \lim_{n \to \infty} \frac{\frac{\pi}{4}}{\frac{\pi}{2}} = \frac{1}{2}. \]

### Final Answer
The limit evaluates to:

\[ \frac{1}{2}. \]