Let the incircle of `DeltaABC` touches the sides BC, CA, AB at `A_(1), B_(1),C_(1)` respectively. The incircle of `DeltaA_(1)B_(1)C_(1)` touches its sides of `B_(1)C_(1), C_(1)A_(1) and A_(1)B_(1)" at " A_(2), B_(2), C_(2)` respectively and so on.
Q. `lim_(n to oo) angleA_(n)=`

Options

A. 0
B. `(pi)/(6)`
C. `(pi)/(4)`
D. `(pi)/(3)`

Answer

To solve the problem, we need to find the limit of the angle \( A_n \) as \( n \) approaches infinity, where \( A_n \) is the angle at vertex A of the triangle formed by the incircles of the triangles formed in each iteration.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The incircle of triangle \( \Delta ABC \) touches the sides at points \( A_1, B_1, C_1 \).
- The incircle of triangle \( \Delta A_1B_1C_1 \) touches its sides at points \( A_2, B_2, C_2 \), and this process continues indefinitely.
- We need to find \( \lim_{n \to \infty} \angle A_n \).

2. **Finding \( \angle A_1 \)**:
- By the properties of tangents to a circle, we know that \( \angle A_1 = \frac{\pi - A}{2} \), where \( A \) is the angle at vertex A in triangle \( ABC \).

3. **Finding \( \angle A_2 \)**:
- For the second triangle \( \Delta A_1B_1C_1 \), we can use the same reasoning:
\[ \angle A_2 = \frac{\pi - \angle A_1}{2} = \frac{\pi - \frac{\pi - A}{2}}{2} = \frac{2\pi - \pi + A}{4} = \frac{\pi + A}{4} \]

4. **Finding \( \angle A_3 \)**:
- Continuing this process, we find:
\[ \angle A_3 = \frac{\pi - \angle A_2}{2} = \frac{\pi - \frac{\pi + A}{4}}{2} = \frac{4\pi - \pi - A}{8} = \frac{3\pi - A}{8} \]

5. **Generalizing \( \angle A_n \)**:
- We can observe a pattern and generalize:
\[ \angle A_n = \frac{(2^n - 1)\pi + (-1)^{n-1} A}{2^n} \]
- This can be derived from the recursive relationship established in the previous steps.

6. **Taking the Limit**:
- Now we take the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} \angle A_n = \lim_{n \to \infty} \left( \frac{(2^n - 1)\pi + (-1)^{n-1} A}{2^n} \right) \]
- As \( n \to \infty \), the term \( \frac{(2^n - 1)\pi}{2^n} \) approaches \( \pi \) and the term \( \frac{(-1)^{n-1} A}{2^n} \) approaches \( 0 \).

7. **Final Result**:
- Thus, we conclude:
\[ \lim_{n \to \infty} \angle A_n = \pi \]

### Conclusion:
The final answer is:
\[ \lim_{n \to \infty} \angle A_n = \pi \]