A function $f: R \to  R$ satisfies $f (x + 1) = f (x) + 1$ for all $x$. Given $a \in R$, define the sequence $(x_n)$ recursively by $x_0 = a$ and $x_{n+1} = f (x_n)$ for $n \ge  0$. Suppose that, for some positive integer m, the difference $x_m - x_0 = k$ is an integer. Prove that the limit $\lim_{n\to \infty}\frac{x_n}{n}$  exists and determine its value.
1. **Claim**: The limit \(\lim_{n \to \infty} \frac{x_n}{n}\) exists and is equal to \(\frac{k}{m}\).

2. **Given**: The function \(f: \mathbb{R} \to \mathbb{R}\) satisfies \(f(x + 1) = f(x) + 1\) for all \(x\). The sequence \((x_n)\) is defined recursively by \(x_0 = a\) and \(x_{n+1} = f(x_n)\) for \(n \geq 0\). For some positive integer \(m\), the difference \(x_m - x_0 = k\) is an integer.

3. **Step 1**: Show that \(f(x + n) = f(x) + n\) for all integers \(n\).
   - By induction on \(n\):
     - Base case (\(n = 1\)): Given \(f(x + 1) = f(x) + 1\).
     - Inductive step: Assume \(f(x + n) = f(x) + n\) holds for some integer \(n\). Then,
       \[
       f(x + (n + 1)) = f((x + n) + 1) = f(x + n) + 1 = (f(x) + n) + 1 = f(x) + (n + 1).
       \]
     - By induction, \(f(x + n) = f(x) + n\) holds for all integers \(n\).

4. **Step 2**: Analyze the sequence \((x_n)\).
   - Given \(x_{n+1} = f(x_n)\), we can write:
     \[
     x_1 = f(x_0), \quad x_2 = f(x_1) = f(f(x_0)), \quad \ldots, \quad x_n = f(x_{n-1}).
     \]
   - Using the property \(f(x + n) = f(x) + n\), we have:
     \[
     x_{m+1} = f(x_m) = f(x_0 + k) = f(x_0) + k = x_1 + k.
     \]
   - Similarly,
     \[
     x_{m+2} = f(x_{m+1}) = f(x_1 + k) = f(x_1) + k = x_2 + k.
     \]
   - By induction, for any integer \(t\),
     \[
     x_{tm + r} = x_r + tk,
     \]
     where \(n = tm + r\) and \(0 \leq r \leq m - 1\).

5. **Step 3**: Determine the limit \(\lim_{n \to \infty} \frac{x_n}{n}\).
   - For \(n = tm + r\),
     \[
     \frac{x_n}{n} = \frac{x_r + tk}{tm + r}.
     \]
   - As \(n \to \infty\), \(t \to \infty\) and \(r\) remains bounded (since \(0 \leq r \leq m - 1\)). Therefore,
     \[
     \frac{x_r}{n} \to 0 \quad \text{and} \quad \frac{tk}{tm + r} \to \frac{tk}{tm} = \frac{k}{m}.
     \]
   - Hence,
     \[
     \lim_{n \to \infty} \frac{x_n}{n} = \frac{k}{m}.
     \]

\(\blacksquare\)

The final answer is \( \boxed{ \frac{k}{m} } \).