Let $f(x)$ be the function defined for $x\geq 0$ which satisfies the following conditions.
(a) $f(x)=\begin{cases}x \ \ \ \ \ \ \ \ ( 0\leq x<1) \\ 2-x \ \ \ (1\leq x <2) \end{cases}$ 

(b) $f(x+2n)=f(x) \ (n=1,2,\cdots)$

Find $\lim_{n\to\infty}\int_{0}^{2n}f(x)e^{-x}\ dx.$
1. **Understanding the function \( f(x) \):**
   The function \( f(x) \) is defined piecewise for \( 0 \leq x < 2 \) and is periodic with period 2. Specifically,
   \[
   f(x) = \begin{cases} 
   x & \text{if } 0 \leq x < 1, \\
   2 - x & \text{if } 1 \leq x < 2.
   \end{cases}
   \]
   Additionally, \( f(x + 2n) = f(x) \) for \( n = 1, 2, \ldots \).

2. **Expressing the integral in terms of the periodicity:**
   Given the periodicity of \( f(x) \), we can break the integral from \( 0 \) to \( 2n \) into \( n \) intervals of length 2:
   \[
   \int_{0}^{2n} f(x) e^{-x} \, dx = \sum_{k=0}^{n-1} \int_{2k}^{2k+2} f(x) e^{-x} \, dx.
   \]
   By the periodicity of \( f(x) \), we have \( f(x + 2k) = f(x) \). Therefore, we can shift the limits of integration:
   \[
   \int_{2k}^{2k+2} f(x) e^{-x} \, dx = \int_{0}^{2} f(x) e^{-(x+2k)} \, dx = e^{-2k} \int_{0}^{2} f(x) e^{-x} \, dx.
   \]

3. **Evaluating the integral over one period:**
   We need to evaluate \( \int_{0}^{2} f(x) e^{-x} \, dx \):
   \[
   \int_{0}^{2} f(x) e^{-x} \, dx = \int_{0}^{1} x e^{-x} \, dx + \int_{1}^{2} (2 - x) e^{-x} \, dx.
   \]

4. **Computing the first integral:**
   \[
   \int_{0}^{1} x e^{-x} \, dx.
   \]
   Using integration by parts, let \( u = x \) and \( dv = e^{-x} dx \). Then \( du = dx \) and \( v = -e^{-x} \):
   \[
   \int_{0}^{1} x e^{-x} \, dx = \left. -x e^{-x} \right|_{0}^{1} + \int_{0}^{1} e^{-x} \, dx = -e^{-1} + \left. -e^{-x} \right|_{0}^{1} = -e^{-1} + (1 - e^{-1}) = 1 - 2e^{-1}.
   \]

5. **Computing the second integral:**
   \[
   \int_{1}^{2} (2 - x) e^{-x} \, dx.
   \]
   Again, using integration by parts, let \( u = 2 - x \) and \( dv = e^{-x} dx \). Then \( du = -dx \) and \( v = -e^{-x} \):
   \[
   \int_{1}^{2} (2 - x) e^{-x} \, dx = \left. -(2 - x) e^{-x} \right|_{1}^{2} + \int_{1}^{2} e^{-x} \, dx = \left. -(2 - x) e^{-x} \right|_{1}^{2} + \left. -e^{-x} \right|_{1}^{2}.
   \]
   Evaluating the boundary terms:
   \[
   \left. -(2 - x) e^{-x} \right|_{1}^{2} = -(2 - 2)e^{-2} + (2 - 1)e^{-1} = e^{-1},
   \]
   and
   \[
   \left. -e^{-x} \right|_{1}^{2} = -e^{-2} + e^{-1}.
   \]
   Therefore,
   \[
   \int_{1}^{2} (2 - x) e^{-x} \, dx = e^{-1} - e^{-2} + e^{-1} = 2e^{-1} - e^{-2}.
   \]

6. **Combining the integrals:**
   \[
   \int_{0}^{2} f(x) e^{-x} \, dx = (1 - 2e^{-1}) + (2e^{-1} - e^{-2}) = 1 - e^{-2}.
   \]

7. **Summing the series:**
   \[
   \int_{0}^{2n} f(x) e^{-x} \, dx = \sum_{k=0}^{n-1} e^{-2k} (1 - e^{-2}).
   \]
   This is a geometric series with the first term \( a = 1 - e^{-2} \) and common ratio \( r = e^{-2} \):
   \[
   \sum_{k=0}^{n-1} e^{-2k} (1 - e^{-2}) = (1 - e^{-2}) \sum_{k=0}^{n-1} e^{-2k} = (1 - e^{-2}) \frac{1 - e^{-2n}}{1 - e^{-2}} = 1 - e^{-2n}.
   \]

8. **Taking the limit as \( n \to \infty \):**
   \[
   \lim_{n \to \infty} \int_{0}^{2n} f(x) e^{-x} \, dx = \lim_{n \to \infty} (1 - e^{-2n}) = 1.
   \]

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