Let $ x$ be an arbitrary real number in $ (0,1)$. For every positive integer $ k$, let $ f_k(x)$ be the number of points $ mx\in [k,k \plus{} 1)$  $ m \equal{} 1,2,...$ 
Show that the sequence $ \sqrt [n]{f_1(x)f_2(x)\cdots f_n(x)}$ is convergent and find its limit.
1. **Define \( f_k(x) \):**
   For a given \( x \in (0,1) \) and a positive integer \( k \), we need to find the number of points \( mx \) in the interval \([k, k+1)\) where \( m \) is a positive integer. This can be expressed as:
   \[
   k \leq mx < k+1
   \]
   Dividing through by \( x \), we get:
   \[
   \frac{k}{x} \leq m < \frac{k+1}{x}
   \]
   Therefore, \( m \) must satisfy:
   \[
   \lceil \frac{k}{x} \rceil \leq m \leq \lfloor \frac{k+1}{x} \rfloor
   \]
   The number of such \( m \) is:
   \[
   f_k(x) = \lfloor \frac{k+1}{x} \rfloor - \lceil \frac{k}{x} \rceil + 1
   \]

2. **Behavior of \( f_k(x) \):**
   - If \( x \) is irrational, \( f_k(x) \) oscillates between \( \lfloor \frac{1}{x} \rfloor \) and \( \lceil \frac{1}{x} \rceil \).
   - If \( x \) is rational, say \( x = \frac{a}{b} \) with \( a \) and \( b \) coprime, then for \( k = nb \) (where \( n \) is an integer), \( f_{nb}(x) \) is:
     \[
     f_{nb}(x) = \lfloor \frac{nb+1}{x} \rfloor - \lfloor \frac{nb}{x} \rfloor = \lfloor \frac{nb+1}{a/b} \rfloor - \lfloor \frac{nb}{a/b} \rfloor = \lfloor n + \frac{1}{x} \rfloor - \lfloor n \rfloor
     \]
     This simplifies to:
     \[
     f_{nb}(x) = \lfloor n + \frac{1}{x} \rfloor - n = \lfloor \frac{1}{x} \rfloor + 1
     \]
     Thus, \( f_k(x) \) still oscillates between \( \lfloor \frac{1}{x} \rfloor \) and \( \lceil \frac{1}{x} \rceil \).

3. **Convergence of the sequence:**
   Consider the sequence \( \sqrt[n]{f_1(x) f_2(x) \cdots f_n(x)} \). Since \( f_k(x) \) oscillates between \( \lfloor \frac{1}{x} \rfloor \) and \( \lceil \frac{1}{x} \rceil \), we can write:
   \[
   \lfloor \frac{1}{x} \rfloor \leq f_k(x) \leq \lceil \frac{1}{x} \rceil
   \]
   Taking the product of \( f_1(x), f_2(x), \ldots, f_n(x) \), we get:
   \[
   \left( \lfloor \frac{1}{x} \rfloor \right)^n \leq f_1(x) f_2(x) \cdots f_n(x) \leq \left( \lceil \frac{1}{x} \rceil \right)^n
   \]
   Taking the \( n \)-th root of each term, we have:
   \[
   \lfloor \frac{1}{x} \rfloor \leq \sqrt[n]{f_1(x) f_2(x) \cdots f_n(x)} \leq \lceil \frac{1}{x} \rceil
   \]
   As \( n \to \infty \), both \( \lfloor \frac{1}{x} \rfloor \) and \( \lceil \frac{1}{x} \rceil \) approach \( \frac{1}{x} \). Therefore, by the Squeeze Theorem, the sequence \( \sqrt[n]{f_1(x) f_2(x) \cdots f_n(x)} \) converges to \( \frac{1}{x} \).

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1}{x} } \)