Let  $({{x}_{n}}),({{y}_{n}})$ be two positive sequences defined by  ${{x}_{1}}=1,{{y}_{1}}=\sqrt{3}$ and 
\[ \begin{cases}  {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\   x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \] for all  $n=1,2,3,\ldots$.
Prove that they are converges and find their limits.
1. **Base Case Verification:**
   We start by verifying the base case for \( n = 1 \). Given \( x_1 = 1 \) and \( y_1 = \sqrt{3} \), we need to check if \( x_1^2 + y_1^2 = 4 \):
   \[
   x_1^2 + y_1^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4
   \]
   Thus, the base case holds.

2. **Inductive Step:**
   Assume that for some \( n = t \), the relation \( x_t^2 + y_t^2 = 4 \) holds. We need to show that \( x_{t+1}^2 + y_{t+1}^2 = 4 \).

3. **Using the Recurrence Relations:**
   From the given recurrence relations:
   \[
   x_{n+1} y_{n+1} = x_n \quad \text{and} \quad x_{n+1}^2 + y_n = 2
   \]
   We can express \( x_{t+1} \) and \( y_{t+1} \) in terms of \( x_t \) and \( y_t \):
   \[
   x_{t+1}^2 = 2 - y_t
   \]
   \[
   y_{t+1} = \frac{x_t}{x_{t+1}}
   \]

4. **Substituting \( y_{t+1} \):**
   Using the inductive hypothesis \( x_t^2 + y_t^2 = 4 \), we substitute \( y_{t+1} \):
   \[
   y_{t+1}^2 = \left( \frac{x_t}{x_{t+1}} \right)^2 = \frac{x_t^2}{x_{t+1}^2}
   \]
   Substituting \( x_{t+1}^2 = 2 - y_t \):
   \[
   y_{t+1}^2 = \frac{x_t^2}{2 - y_t}
   \]

5. **Simplifying \( y_{t+1}^2 \):**
   Using the inductive hypothesis \( x_t^2 = 4 - y_t^2 \):
   \[
   y_{t+1}^2 = \frac{4 - y_t^2}{2 - y_t}
   \]

6. **Summing \( x_{t+1}^2 \) and \( y_{t+1}^2 \):**
   \[
   x_{t+1}^2 + y_{t+1}^2 = (2 - y_t) + \frac{4 - y_t^2}{2 - y_t}
   \]
   Simplify the second term:
   \[
   \frac{4 - y_t^2}{2 - y_t} = \frac{(2 - y_t)(2 + y_t)}{2 - y_t} = 2 + y_t
   \]
   Therefore:
   \[
   x_{t+1}^2 + y_{t+1}^2 = (2 - y_t) + (2 + y_t) = 4
   \]
   This completes the inductive step.

7. **Conclusion:**
   By induction, \( x_n^2 + y_n^2 = 4 \) for all \( n \).

8. **Finding the Limits:**
   Let \( x_n = 2 \sin a_n \) and \( y_n = 2 \cos a_n \) where \( a_n \in (0, \pi/2) \). From the recurrence relations:
   \[
   x_{n+1}^2 = 4 \sin^2 a_{n+1} = 2(1 - \cos a_n)
   \]
   \[
   2 \sin^2 a_{n+1} = 1 - \cos a_n
   \]
   Since the sequences are positive, we have:
   \[
   a_{n+1} = \frac{a_n}{2}
   \]
   As \( n \to \infty \), \( a_n \to 0 \), so \( x_n \to 0 \) and \( y_n \to 2 \).

The final answer is \( \boxed{ \lim_{n \to \infty} x_n = 0 } \) and \( \lim_{n \to \infty} y_n = 2 \).