Determine all ordered pairs of positive real numbers $(a, b)$ such that every sequence $(x_{n})$ satisfying $\lim_{n \rightarrow \infty}{(ax_{n+1} - bx_{n})} = 0$ must have $\lim_{n \rightarrow \infty} x_n = 0$.
To determine all ordered pairs of positive real numbers \((a, b)\) such that every sequence \((x_n)\) satisfying \(\lim_{n \rightarrow \infty}{(ax_{n+1} - bx_{n})} = 0\) must have \(\lim_{n \rightarrow \infty} x_n = 0\), we proceed as follows:

1. **Case \(a = 0\):**
   - If \(a = 0\), the condition becomes \(\lim_{n \rightarrow \infty} (-bx_n) = 0\).
   - Since \(b \neq 0\), this implies \(\lim_{n \rightarrow \infty} x_n = 0\).
   - Therefore, \(a = 0\) and \(b \neq 0\) satisfy the given condition.

2. **Case \(a \neq 0\):**
   - Let \(c = \frac{b}{a}\). The given condition becomes \(\lim_{n \rightarrow \infty} (ax_{n+1} - bx_n) = 0\), which simplifies to:
     \[
     \lim_{n \rightarrow \infty} (x_{n+1} - cx_n) = 0
     \]
   - This implies:
     \[
     x_{n+1} = cx_n + \epsilon_n
     \]
     where \(\epsilon_n \to 0\) as \(n \to \infty\).

3. **Analyzing the sequence:**
   - From the recurrence relation, we have:
     \[
     x_{n+1} = \sum_{j=0}^{n-1} c^j \epsilon_{n-j} + c^n x_1
     \]
   - We need to analyze the behavior of \(x_n\) based on the value of \(c\).

4. **Case \(c \ge 1\):**
   - If \(c \ge 1\) and we choose \(\epsilon_j = \frac{1}{j}\) and \(x_1 = 0\), then:
     \[
     x_{n+1} > \sum_{j=1}^n \frac{1}{j}
     \]
   - This implies \(x_n \to \infty\), which contradicts the requirement that \(\lim_{n \to \infty} x_n = 0\).

5. **Case \(c \le -1\):**
   - The analysis is similar to the case \(c \ge 1\), leading to \(x_n \to \infty\).

6. **Case \(|c| < 1\):**
   - We will prove that \(x_n \to 0\).
   - Since \(\epsilon_n \to 0\), there exists a constant \(A > 0\) such that \(|\epsilon_n| < A\) for all \(n\).
   - Fix \(\epsilon > 0\). There exists \(N\) such that \(|\epsilon_j| < \epsilon\) for \(j > N\).
   - For \(n > N\), we estimate \(x_n\) as follows:
     \[
     |x_{n+1}| < \sum_{j=0}^{n-N} |c|^j |\epsilon_{n-j}| + \sum_{j=n-N+1}^{n-1} |c|^j |\epsilon_{n-j}| + |c|^n |x_1|
     \]
     \[
     < \epsilon \sum_{j=0}^{n-N} |c|^j + A \sum_{j=n-N+1}^{n-1} |c|^j + |c|^n |x_1|
     \]
     \[
     < \epsilon \frac{1 - |c|^{n-N+1}}{1 - |c|} + A \frac{|c|^{n-N+1} - |c|^n}{1 - |c|} + |c|^n |x_1|
     \]
     \[
     < \epsilon \frac{1}{1 - |c|} + A \frac{|c|^{n-N+1}}{1 - |c|} + |c|^n |x_1|
     \]
   - Since \(|c| < 1\), as \(n \to \infty\), \(|c|^n \to 0\). Therefore, for large \(n\), \(|x_n| < 2 \epsilon \frac{1}{1 - |c|}\), implying \(\lim_{n \to \infty} x_n = 0\).

Conclusion:
- The pairs \((a, b)\) that satisfy the given condition are:
  1. \(a = 0\), \(b \neq 0\)
  2. \(a \neq 0\), \(|\frac{b}{a}| < 1\)

The final answer is \( \boxed{ (a, b) } \) such that \(a = 0\) and \(b \neq 0\) or \(a \neq 0\) and \(|\frac{b}{a}| < 1\).