Answer
$\lim_{n \to \infty} x_n = 2$
Work Step by Step
We have the sequence \(\{x_k\}\) defined by
\[
x_0 = 0, \quad\text{and for } k \ge 1:\; x_k = \sqrt{2 + x_{k-1}}.
\]
We wish to find
\[
\lim_{n \to \infty} x_n,
\]
assuming this limit exists.
---
## 1. Assume the Limit Exists
Let
\[
L \;=\; \lim_{n \to \infty} x_n.
\]
Because each \(x_k\) is defined in terms of the previous term, for large \(n\) we have
\[
x_{n+1} \;=\; \sqrt{2 + x_n}.
\]
If the limit \(L\) exists, then \(x_n \to L\) and \(x_{n+1} \to L\) as \(n \to \infty\). Hence, in the limit,
\[
L = \sqrt{2 + L}.
\]
---
## 2. Solve the Equation \(L = \sqrt{2 + L}\)
Square both sides:
\[
L^2 = 2 + L,
\]
or
\[
L^2 - L - 2 = 0.
\]
This is a standard quadratic equation. Solve via the quadratic formula:
\[
L = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}.
\]
Thus, the two possible solutions are
\[
L = 2 \quad \text{or} \quad L = -1.
\]
---
## 3. Choose the Nonnegative Root
Since \(x_0=0\) and each subsequent term \(x_k = \sqrt{2 + x_{k-1}}\) is clearly nonnegative (the square root is always nonnegative, and the sequence is increasing or at least nondecreasing), the limit \(L\) must also be nonnegative. Hence we discard \(L=-1\).
Therefore,
\[
L = 2.
\]
---
## 4. Conclusion
Provided the limit exists (which one can also justify by showing the sequence is monotone increasing and bounded above by 2), we have
\[
\boxed{\lim_{n \to \infty} x_n = 2}.
\]
