[b]6.[/b] Let $H_{n}(x)$ be the [i]n[/i]th Hermite polynomial. Find
$ \lim_{n \to \infty } (\frac{y}{2n})^{n} H_{n}(\frac{n}{y})$
For an arbitrary real y. [b](S.5)[/b]

$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}\left(e^{{-x^2}}\right)$
1. **Definition and Base Case:**
   The Hermite polynomial \( H_n(x) \) is defined as:
   \[
   H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} \left( e^{-x^2} \right)
   \]
   For the base case, \( H_0(x) = 1 \).

2. **Recurrence Relation:**
   We use the recurrence relation for Hermite polynomials:
   \[
   H_{n+1}(x) = 2xH_n(x) - H'_n(x)
   \]
   This can be derived from the definition of Hermite polynomials.

3. **Leading Term Analysis:**
   We are interested in the leading term of \( H_n(x) \). We observe that each time we transition from \( H_n(x) \) to \( H_{n+1}(x) \), the highest-degree term is multiplied by \( 2x \). Starting with \( H_0(x) = 1 \), the leading term of \( H_n(x) \) is \( (2x)^n \).

4. **Limit Calculation:**
   We need to find:
   \[
   \lim_{n \to \infty} \left( \frac{y}{2n} \right)^n H_n \left( \frac{n}{y} \right)
   \]
   Substituting the leading term of \( H_n(x) \):
   \[
   H_n \left( \frac{n}{y} \right) \approx \left( 2 \cdot \frac{n}{y} \right)^n = \left( \frac{2n}{y} \right)^n
   \]
   Therefore, the expression becomes:
   \[
   \left( \frac{y}{2n} \right)^n \left( \frac{2n}{y} \right)^n = \left( \frac{y}{2n} \cdot \frac{2n}{y} \right)^n = 1^n = 1
   \]

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