Define the sequence $x_1, x_2, ...$ inductively by $x_1 = \sqrt{5}$ and $x_{n+1} = x_n^2 - 2$ for each $n \geq 1$. Compute
$\lim_{n \to \infty} \frac{x_1 \cdot x_2 \cdot x_3 \cdot ... \cdot x_n}{x_{n+1}}$.
1. **Define the sequence and the limit to be computed:**
   The sequence \( \{x_n\} \) is defined by:
   \[
   x_1 = \sqrt{5}, \quad x_{n+1} = x_n^2 - 2 \quad \text{for} \quad n \geq 1.
   \]
   We need to compute:
   \[
   \lim_{n \to \infty} \frac{x_1 \cdot x_2 \cdot x_3 \cdot \ldots \cdot x_n}{x_{n+1}}.
   \]

2. **Lemma and Induction Hypothesis:**
   We claim that for all \( n \geq 1 \):
   \[
   \frac{x_1 x_2 \cdots x_n}{x_{n+1}} = \sqrt{\frac{x_{n+1}^2 - 4}{x_{n+1}^2}}.
   \]
   We will prove this by induction.

3. **Base Case:**
   For \( n = 1 \):
   \[
   \frac{x_1}{x_2} = \frac{\sqrt{5}}{(\sqrt{5})^2 - 2} = \frac{\sqrt{5}}{5 - 2} = \frac{\sqrt{5}}{3}.
   \]
   Now, check the lemma for \( n = 1 \):
   \[
   \sqrt{\frac{x_2^2 - 4}{x_2^2}} = \sqrt{\frac{(5 - 2)^2 - 4}{(5 - 2)^2}} = \sqrt{\frac{9 - 4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}.
   \]
   Thus, the base case holds.

4. **Inductive Step:**
   Assume the lemma holds for some \( k \geq 1 \):
   \[
   \frac{x_1 x_2 \cdots x_k}{x_{k+1}} = \sqrt{\frac{x_{k+1}^2 - 4}{x_{k+1}^2}}.
   \]
   We need to show it holds for \( k+1 \):
   \[
   \frac{x_1 x_2 \cdots x_{k+1}}{x_{k+2}} = \frac{x_1 x_2 \cdots x_k}{x_{k+1}} \cdot \frac{x_{k+1}}{x_{k+2}}.
   \]
   Using the induction hypothesis:
   \[
   \frac{x_1 x_2 \cdots x_{k+1}}{x_{k+2}} = \sqrt{\frac{x_{k+1}^2 - 4}{x_{k+1}^2}} \cdot \frac{x_{k+1}}{x_{k+2}}.
   \]
   Substitute \( x_{k+2} = x_{k+1}^2 - 2 \):
   \[
   \frac{x_1 x_2 \cdots x_{k+1}}{x_{k+2}} = \sqrt{\frac{x_{k+1}^2 - 4}{x_{k+1}^2}} \cdot \frac{x_{k+1}}{x_{k+1}^2 - 2}.
   \]
   Simplify the expression:
   \[
   \frac{x_1 x_2 \cdots x_{k+1}}{x_{k+2}} = \sqrt{\frac{x_{k+1}^2 - 4}{x_{k+1}^2}} \cdot \frac{x_{k+1}}{x_{k+1}^2 - 2} = \sqrt{\frac{(x_{k+1}^2 - 4) x_{k+1}^2}{(x_{k+1}^2 - 2)^2}}.
   \]
   Since \( x_{k+2} = x_{k+1}^2 - 2 \):
   \[
   \frac{x_1 x_2 \cdots x_{k+1}}{x_{k+2}} = \sqrt{\frac{(x_{k+2} + 2)(x_{k+2} - 2)}{x_{k+2}^2}} = \sqrt{\frac{x_{k+2}^2 - 4}{x_{k+2}^2}}.
   \]
   Thus, the inductive step holds.

5. **Conclusion:**
   By induction, the lemma holds for all \( n \geq 1 \):
   \[
   \frac{x_1 x_2 \cdots x_n}{x_{n+1}} = \sqrt{\frac{x_{n+1}^2 - 4}{x_{n+1}^2}}.
   \]

6. **Limit Calculation:**
   \[
   \lim_{n \to \infty} \frac{x_1 \cdot x_2 \cdot x_3 \cdot \ldots \cdot x_n}{x_{n+1}} = \lim_{n \to \infty} \sqrt{\frac{x_{n+1}^2 - 4}{x_{n+1}^2}} = \sqrt{\lim_{n \to \infty} \left(1 - \frac{4}{x_{n+1}^2}\right)}.
   \]
   As \( n \to \infty \), \( x_n \to \infty \), so:
   \[
   \lim_{n \to \infty} \frac{4}{x_{n+1}^2} = 0.
   \]
   Therefore:
   \[
   \lim_{n \to \infty} \sqrt{1 - \frac{4}{x_{n+1}^2}} = \sqrt{1 - 0} = 1.
   \]

The final answer is \(\boxed{1}\).