Let $\{x_n\}_{n=1}^\infty$ be a sequence such that $x_1=25$, $x_n=\operatorname{arctan}(x_{n-1})$. Prove that this sequence has a limit and find it.
1. **Initial Sequence Definition and Properties**:
   - Given the sequence $\{x_n\}_{n=1}^\infty$ with $x_1 = 25$ and $x_n = \arctan(x_{n-1})$ for $n \geq 2$.
   - We need to show that this sequence has a limit and find it.

2. **Monotonicity and Boundedness**:
   - For $x > 0$, we have $0 < \arctan(x) < x$. This is because $\arctan(x)$ is the integral $\arctan(x) = \int_0^x \frac{dt}{t^2 + 1}$, and since $t^2 + 1 > 1$ for all $t \geq 0$, we have $\arctan(x) < \int_0^x dt = x$.
   - Therefore, $x_n > 0$ for all $n$ and $x_n$ is strictly decreasing.

3. **Convergence**:
   - Since $x_n$ is strictly decreasing and bounded below by 0, it must converge to some limit $a \geq 0$.

4. **Fixed Point Analysis**:
   - If $x_n \to a$, then taking the limit on both sides of the recurrence relation $x_n = \arctan(x_{n-1})$, we get:
     \[
     a = \arctan(a)
     \]
   - The function $\arctan(x)$ has a unique fixed point at $a = 0$ because $\arctan(x) = x$ only when $x = 0$.

5. **Conclusion**:
   - Therefore, the limit of the sequence $\{x_n\}_{n=1}^\infty$ is $a = 0$.

The final answer is $\boxed{0}$.