The function $f_n (x)\ (n=1,2,\cdots)$ is defined  as follows.

\[f_1 (x)=x,\ f_{n+1}(x)=2x^{n+1}-x^n+\frac{1}{2}\int_0^1 f_n(t)\ dt\ \ (n=1,2,\cdots)\]

Evaluate 

\[\lim_{n\to\infty} f_n \left(1+\frac{1}{2n}\right)\]
1. **Define the sequence \( A_n \):**
   \[
   A_n = \int_0^1 f_n(t) \, dt
   \]
   Given \( f_1(x) = x \), we have:
   \[
   A_1 = \int_0^1 t \, dt = \left[ \frac{t^2}{2} \right]_0^1 = \frac{1}{2}
   \]

2. **Calculate \( A_2 \):**
   Using the recurrence relation \( f_{n+1}(x) = 2x^{n+1} - x^n + \frac{1}{2} A_n \), we find:
   \[
   f_2(x) = 2x^2 - x + \frac{1}{2} \int_0^1 t \, dt = 2x^2 - x + \frac{1}{4}
   \]
   Then,
   \[
   A_2 = \int_0^1 (2t^2 - t + \frac{1}{4}) \, dt = \left[ \frac{2t^3}{3} - \frac{t^2}{2} + \frac{t}{4} \right]_0^1 = \frac{2}{3} - \frac{1}{2} + \frac{1}{4} = \frac{8}{12} - \frac{6}{12} + \frac{3}{12} = \frac{5}{12}
   \]

3. **Prove \( A_{n+1} < A_n \) by induction:**
   - Base case: \( A_2 < A_1 \) is already shown.
   - Inductive step: Assume \( A_n < A_{n-1} \) for some \( n \geq 2 \). We need to show \( A_{n+1} < A_n \).

     Using the recurrence relation:
     \[
     A_{n+1} = \int_0^1 f_{n+1}(t) \, dt = \int_0^1 \left( 2t^{n+1} - t^n + \frac{1}{2} A_n \right) \, dt
     \]
     \[
     A_{n+1} = \left[ \frac{2t^{n+2}}{n+2} - \frac{t^{n+1}}{n+1} \right]_0^1 + \frac{1}{2} A_n = \frac{2}{n+2} - \frac{1}{n+1} + \frac{1}{2} A_n
     \]
     We need to show:
     \[
     \frac{2}{n+2} - \frac{1}{n+1} + \frac{1}{2} A_n < A_n
     \]
     Simplifying:
     \[
     \frac{2}{n+2} - \frac{1}{n+1} < \frac{1}{2} A_n
     \]
     Since \( A_n \) is positive and decreasing, the inequality holds for \( n \geq 2 \).

4. **Show \( \lim_{n \to \infty} A_n = 0 \):**
   Since \( A_n \) is a decreasing sequence bounded below by 0, it converges to some limit \( C \geq 0 \). From the recurrence relation:
   \[
   A_{n+1} = \frac{2}{n+2} - \frac{1}{n+1} + \frac{1}{2} A_n
   \]
   Taking the limit as \( n \to \infty \):
   \[
   C = \lim_{n \to \infty} A_{n+1} = \lim_{n \to \infty} \left( \frac{2}{n+2} - \frac{1}{n+1} + \frac{1}{2} A_n \right) = 0 + 0 + \frac{1}{2} C
   \]
   \[
   C = \frac{1}{2} C \implies C = 0
   \]

5. **Evaluate \( \lim_{n \to \infty} f_n \left(1 + \frac{1}{2n}\right) \):**
   Using the recurrence relation:
   \[
   f_{n+1}(x) = 2x^{n+1} - x^n + \frac{1}{2} A_n
   \]
   As \( n \to \infty \), \( A_n \to 0 \), so:
   \[
   f_{n+1}(x) \approx 2x^{n+1} - x^n
   \]
   For \( x = 1 + \frac{1}{2n} \):
   \[
   f_{n+1}\left(1 + \frac{1}{2n}\right) \approx \left(1 + \frac{1}{2n}\right)^n \left(2 \left(1 + \frac{1}{2n}\right) - 1\right)
   \]
   \[
   \approx e^{1/2} \left(1 + \frac{1}{n}\right) \approx e^{1/2}
   \]

The final answer is \(\boxed{e^{1/2}}\).