Let $E(n)$ denote the largest integer $k$ such that $5^k$ divides $1^{1}\cdot 2^{2} \cdot 3^{3} \cdot \ldots \cdot n^{n}.$ Calculate
$$\lim_{n\to \infty} \frac{E(n)}{n^2 }.$$
To solve the problem, we need to determine the largest integer \( k \) such that \( 5^k \) divides the product \( 1^1 \cdot 2^2 \cdot 3^3 \cdot \ldots \cdot n^n \). We denote this integer by \( E(n) \). We then need to find the limit:
\[
\lim_{n \to \infty} \frac{E(n)}{n^2}.
\]

1. **Determine the exponent of 5 in the product**:
   The exponent of 5 in the product \( 1^1 \cdot 2^2 \cdot 3^3 \cdot \ldots \cdot n^n \) is given by the sum of the exponents of 5 in each term \( k^k \) for \( k = 1, 2, \ldots, n \).

2. **Exponent of 5 in \( k^k \)**:
   For a given \( k \), the exponent of 5 in \( k^k \) is \( k \) times the exponent of 5 in \( k \). Let \( v_5(k) \) denote the exponent of 5 in \( k \). Then the exponent of 5 in \( k^k \) is \( k \cdot v_5(k) \).

3. **Sum the exponents**:
   We need to sum \( k \cdot v_5(k) \) for \( k = 1 \) to \( n \):
   \[
   E(n) = \sum_{k=1}^n k \cdot v_5(k).
   \]

4. **Calculate \( v_5(k) \)**:
   The function \( v_5(k) \) counts the number of times 5 divides \( k \). For example, \( v_5(25) = 2 \) because \( 25 = 5^2 \).

5. **Group terms by multiples of 5**:
   To simplify the sum, we group terms by multiples of 5. For each multiple of 5, we need to account for the additional factors of 5:
   \[
   E(n) = \sum_{m=1}^{\infty} \left( \sum_{k=1}^{\left\lfloor \frac{n}{5^m} \right\rfloor} 5^m k \right).
   \]
   This sum accounts for the contributions from \( 5, 25, 125, \ldots \).

6. **Approximate the sum**:
   For large \( n \), the inner sum can be approximated by:
   \[
   \sum_{k=1}^{\left\lfloor \frac{n}{5^m} \right\rfloor} 5^m k \approx 5^m \cdot \frac{\left\lfloor \frac{n}{5^m} \right\rfloor (\left\lfloor \frac{n}{5^m} \right\rfloor + 1)}{2}.
   \]
   For large \( n \), this approximates to:
   \[
   \approx 5^m \cdot \frac{\left( \frac{n}{5^m} \right)^2}{2} = \frac{n^2}{2 \cdot 5^m}.
   \]

7. **Sum over all powers of 5**:
   Summing over all powers of 5, we get:
   \[
   E(n) \approx \sum_{m=1}^{\infty} \frac{n^2}{2 \cdot 5^m}.
   \]
   This is a geometric series with the first term \( \frac{n^2}{10} \) and common ratio \( \frac{1}{5} \):
   \[
   E(n) \approx \frac{n^2}{10} \sum_{m=0}^{\infty} \left( \frac{1}{5} \right)^m = \frac{n^2}{10} \cdot \frac{1}{1 - \frac{1}{5}} = \frac{n^2}{10} \cdot \frac{5}{4} = \frac{n^2}{8}.
   \]

8. **Calculate the limit**:
   \[
   \lim_{n \to \infty} \frac{E(n)}{n^2} = \lim_{n \to \infty} \frac{\frac{n^2}{8}}{n^2} = \frac{1}{8}.
   \]

The final answer is \(\boxed{\frac{1}{8}}\).