Let $a\in \mathbb{R}_+$ and define the sequence of real numbers $(x_n)_n$ by $x_1=a$ and $x_{n+1}=\left|x_n-\frac{1}{n}\right|,\ n\ge 1$. Prove that the sequence is convergent and find it's limit.
1. **Lemma 1**: If, for some \( n \) we have \( x_n < \frac{1}{n} \), then for all \( m > n \) we have \( x_m < \frac{1}{n} \).

   **Proof**: 
   - Assume \( x_n < \frac{1}{n} \).
   - For \( m = n \), we have \( x_{n+1} = \left| x_n - \frac{1}{n} \right| \).
   - Since \( 0 < x_n < \frac{1}{n} \), it follows that \( 0 < \frac{1}{n} - x_n < \frac{1}{n} \).
   - Therefore, \( x_{n+1} = \frac{1}{n} - x_n < \frac{1}{n} \).
   - By induction, if \( x_k < \frac{1}{n} \) for some \( k \geq n \), then \( x_{k+1} = \left| x_k - \frac{1}{k} \right| < \frac{1}{n} \) because \( 0 < x_k < \frac{1}{n} \) and \( 0 < \frac{1}{k} < \frac{1}{n} \).
   - Hence, \( x_m < \frac{1}{n} \) for all \( m > n \).

2. **Lemma 2**: For every \( n \), there is an \( m \) such that \( x_m < \frac{1}{n} \).

   **Proof**:
   - Fix \( n \). Suppose that for every \( m \), \( x_m \geq \frac{1}{n} \).
   - In particular, for \( m = n, n+1, \ldots \), we have \( x_{n+k} \geq \frac{1}{n} \geq \frac{1}{n+k} \).
   - Therefore, \( x_{n+k+1} = x_{n+k} - \frac{1}{n+k} \).
   - Recursively, for every \( k \), we have:
     \[
     x_{n+k} = x_n - \left( \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{n+k-1} \right)
     \]
   - Since the harmonic series diverges, for large enough \( k \), the right-hand side becomes negative, which contradicts the assumption that \( x_m \geq \frac{1}{n} \).
   - Hence, there must exist some \( m \) such that \( x_m < \frac{1}{n} \).

3. Combining Lemma 1 and Lemma 2:
   - Lemma 2 ensures that for any \( n \), there exists an \( m \) such that \( x_m < \frac{1}{n} \).
   - Lemma 1 then ensures that for all \( k \geq m \), \( x_k < \frac{1}{n} \).
   - This implies that for any given \( \epsilon > 0 \), there exists an \( N \) such that for all \( n \geq N \), \( x_n < \epsilon \).
   - Therefore, \( \lim_{n \to \infty} x_n = 0 \).

The final answer is \( \boxed{ 0 } \).