For $a>1$, let $f(a)=\frac{1}{2}\int_0^1 |ax^n-1|dx+\frac{1}{2}\ (n=1,2,\cdots)$ and let $b_n$ be the minimum value of $f(a)$ at $a>1$.

Evaluate 

\[\lim_{m\to\infty} b_m\cdot b_{m+1}\cdot \cdots\cdots b_{2m}\ (m=1,2,3,\cdots)\]
1. Given the function \( f(a) = \frac{1}{2} \int_0^1 |ax^n - 1| \, dx + \frac{1}{2} \), we need to find the minimum value of \( f(a) \) for \( a > 1 \).
2. To evaluate \( f(a) \), we first consider the integral \( \int_0^1 |ax^n - 1| \, dx \). This integral can be split into two parts based on the value of \( x \):
   - For \( x \in [0, a^{-1/n}] \), \( ax^n \leq 1 \) and \( |ax^n - 1| = 1 - ax^n \).
   - For \( x \in [a^{-1/n}, 1] \), \( ax^n \geq 1 \) and \( |ax^n - 1| = ax^n - 1 \).
3. Therefore, the integral can be written as:
   \[
   \int_0^1 |ax^n - 1| \, dx = \int_0^{a^{-1/n}} (1 - ax^n) \, dx + \int_{a^{-1/n}}^1 (ax^n - 1) \, dx
   \]
4. Evaluating these integrals separately:
   \[
   \int_0^{a^{-1/n}} (1 - ax^n) \, dx = \int_0^{a^{-1/n}} 1 \, dx - \int_0^{a^{-1/n}} ax^n \, dx = a^{-1/n} - \frac{a}{n+1} \left( x^{n+1} \bigg|_0^{a^{-1/n}} \right) = a^{-1/n} - \frac{a}{n+1} \cdot a^{-(n+1)/n}
   \]
   Simplifying, we get:
   \[
   a^{-1/n} - \frac{a}{n+1} \cdot a^{-1-1/n} = a^{-1/n} - \frac{1}{(n+1)a^{1/n}}
   \]
5. For the second part:
   \[
   \int_{a^{-1/n}}^1 (ax^n - 1) \, dx = \int_{a^{-1/n}}^1 ax^n \, dx - \int_{a^{-1/n}}^1 1 \, dx = \frac{a}{n+1} \left( x^{n+1} \bigg|_{a^{-1/n}}^1 \right) - (1 - a^{-1/n})
   \]
   Simplifying, we get:
   \[
   \frac{a}{n+1} (1 - a^{-1}) - (1 - a^{-1/n}) = \frac{a}{n+1} - \frac{1}{(n+1)a} - 1 + a^{-1/n}
   \]
6. Combining both parts, we have:
   \[
   \int_0^1 |ax^n - 1| \, dx = a^{-1/n} - \frac{1}{(n+1)a^{1/n}} + \frac{a}{n+1} - \frac{1}{(n+1)a} - 1 + a^{-1/n}
   \]
7. Simplifying further, we get:
   \[
   \int_0^1 |ax^n - 1| \, dx = 2a^{-1/n} - \frac{1}{(n+1)a^{1/n}} + \frac{a}{n+1} - \frac{1}{(n+1)a} - 1
   \]
8. Therefore, the function \( f(a) \) becomes:
   \[
   f(a) = \frac{1}{2} \left( 2a^{-1/n} - \frac{1}{(n+1)a^{1/n}} + \frac{a}{n+1} - \frac{1}{(n+1)a} - 1 \right) + \frac{1}{2}
   \]
9. To find the minimum value of \( f(a) \), we need to minimize the expression inside the integral. As \( n \to \infty \), the terms involving \( a^{-1/n} \) and \( a^{1/n} \) approach 1, and the dominant term becomes \( \frac{a}{n+1} \).
10. Hence, the minimum value \( b_n \) of \( f(a) \) as \( n \to \infty \) is approximately:
    \[
    b_n \approx \frac{1}{2} \left( 1 + \left| \frac{a}{n+1} - 1 \right| \right)
    \]
11. As \( n \to \infty \), \( b_n \) approaches 1. Therefore, the product \( b_m \cdot b_{m+1} \cdot \cdots \cdot b_{2m} \) also approaches 1.

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