For each positive integer $n$, define $H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}.$

(1) Find $H_1(x),\ H_2(x),\ H_3(x)$.

(2) Express $\frac{d}{dx}H_n(x)$ interms of $H_n(x),\ H_{n+1}(x).$ Then prove that $H_n(x)$ is a polynpmial with degree $n$ by induction.

(3) Let $a$ be real number. For $n\geq 3$, express $S_n(a)=\int_0^a xH_n(x)e^{-x^2}dx$ in terms of $H_{n-1}(a),\ H_{n-2}(a),\ H_{n-2}(0)$.

(4) Find $\lim_{a\to\infty} S_6(a)$.
If necessary, you may use $\lim_{x\to\infty}x^ke^{-x^2}=0$ for a positive integer $k$.
### Part (1)
We need to find \( H_1(x), H_2(x), H_3(x) \).

1. **For \( H_1(x) \):**
   \[
   H_1(x) = (-1)^1 e^{x^2} \frac{d}{dx} e^{-x^2}
   \]
   \[
   \frac{d}{dx} e^{-x^2} = -2x e^{-x^2}
   \]
   \[
   H_1(x) = -e^{x^2} \cdot (-2x e^{-x^2}) = 2x
   \]

2. **For \( H_2(x) \):**
   \[
   H_2(x) = (-1)^2 e^{x^2} \frac{d^2}{dx^2} e^{-x^2}
   \]
   \[
   \frac{d}{dx} (-2x e^{-x^2}) = -2 e^{-x^2} + 4x^2 e^{-x^2} = (4x^2 - 2) e^{-x^2}
   \]
   \[
   H_2(x) = e^{x^2} (4x^2 - 2) e^{-x^2} = 4x^2 - 2
   \]

3. **For \( H_3(x) \):**
   \[
   H_3(x) = (-1)^3 e^{x^2} \frac{d^3}{dx^3} e^{-x^2}
   \]
   \[
   \frac{d}{dx} (4x^2 - 2) e^{-x^2} = (8x - 8x^3) e^{-x^2}
   \]
   \[
   \frac{d^2}{dx^2} e^{-x^2} = (8x - 8x^3) e^{-x^2} + (4x^2 - 2) (-2x) e^{-x^2} = (8x - 8x^3 - 8x^3 + 4x^2) e^{-x^2}
   \]
   \[
   H_3(x) = -e^{x^2} (8x^3 - 12x) e^{-x^2} = - (8x^3 - 12x) = 8x^3 - 12x
   \]

### Part (2)
We need to express \(\frac{d}{dx} H_n(x)\) in terms of \(H_n(x)\) and \(H_{n+1}(x)\), and prove by induction that \(H_n(x)\) is a polynomial of degree \(n\).

1. **Expressing \(\frac{d}{dx} H_n(x)\):**
   \[
   \frac{d}{dx} H_n(x) = (-1)^n \left( \frac{d}{dx} \left( e^{x^2} \frac{d^n}{dx^n} e^{-x^2} \right) \right)
   \]
   Using the product rule:
   \[
   \frac{d}{dx} \left( e^{x^2} \frac{d^n}{dx^n} e^{-x^2} \right) = 2x e^{x^2} \frac{d^n}{dx^n} e^{-x^2} + e^{x^2} \frac{d^{n+1}}{dx^{n+1}} e^{-x^2}
   \]
   \[
   \frac{d}{dx} H_n(x) = (-1)^n \left( 2x e^{x^2} \frac{d^n}{dx^n} e^{-x^2} + e^{x^2} \frac{d^{n+1}}{dx^{n+1}} e^{-x^2} \right)
   \]
   \[
   \frac{d}{dx} H_n(x) = 2x H_n(x) - H_{n+1}(x)
   \]

2. **Induction Proof:**
   - **Base Case:** For \(n=1\), \(H_1(x) = 2x\), which is a polynomial of degree 1.
   - **Inductive Step:** Assume \(H_k(x)\) is a polynomial of degree \(k\). Then,
     \[
     H_{k+1}(x) = 2x H_k(x) - \frac{d}{dx} H_k(x)
     \]
     Since \(H_k(x)\) is a polynomial of degree \(k\), \(2x H_k(x)\) is a polynomial of degree \(k+1\) and \(\frac{d}{dx} H_k(x)\) is a polynomial of degree \(k-1\). Thus, \(H_{k+1}(x)\) is a polynomial of degree \(k+1\).

   By induction, \(H_n(x)\) is a polynomial of degree \(n\) for all \(n \in \mathbb{N}^+\).

### Part (3)
We need to express \(S_n(a) = \int_0^a x H_n(x) e^{-x^2} \, dx\) in terms of \(H_{n-1}(a), H_{n-2}(a), H_{n-2}(0)\).

1. **Using integration by parts:**
   \[
   S_n(a) = \int_0^a x H_n(x) e^{-x^2} \, dx
   \]
   Let \(u = x H_n(x)\) and \(dv = e^{-x^2} dx\). Then \(du = (H_n(x) + x H_n'(x)) dx\) and \(v = -\frac{1}{2} e^{-x^2}\).
   \[
   S_n(a) = \left[ -\frac{1}{2} x H_n(x) e^{-x^2} \right]_0^a + \frac{1}{2} \int_0^a (H_n(x) + x H_n'(x)) e^{-x^2} \, dx
   \]
   \[
   S_n(a) = -\frac{1}{2} a H_n(a) e^{-x^2} + \frac{1}{2} \int_0^a H_n(x) e^{-x^2} \, dx + \frac{1}{2} \int_0^a x H_n'(x) e^{-x^2} \, dx
   \]
   Using \(H_n'(x) = 2x H_n(x) - H_{n+1}(x)\):
   \[
   \int_0^a x H_n'(x) e^{-x^2} \, dx = \int_0^a x (2x H_n(x) - H_{n+1}(x)) e^{-x^2} \, dx
   \]
   \[
   = 2 \int_0^a x^2 H_n(x) e^{-x^2} \, dx - \int_0^a x H_{n+1}(x) e^{-x^2} \, dx
   \]
   \[
   S_n(a) = -\frac{1}{2} a H_n(a) e^{-x^2} + \frac{1}{2} \int_0^a H_n(x) e^{-x^2} \, dx + \frac{1}{2} \left( 2 \int_0^a x^2 H_n(x) e^{-x^2} \, dx - \int_0^a x H_{n+1}(x) e^{-x^2} \, dx \right)
   \]
   \[
   S_n(a) = -\frac{1}{2} a H_n(a) e^{-x^2} + \frac{1}{2} \int_0^a H_n(x) e^{-x^2} \, dx + \int_0^a x^2 H_n(x) e^{-x^2} \, dx - \frac{1}{2} \int_0^a x H_{n+1}(x) e^{-x^2} \, dx
   \]

### Part (4)
We need to find \(\lim_{a \to \infty} S_6(a)\).

1. **Using the result from Part (3):**
   \[
   S_6(a) = -e^{-a^2} a H_5(a) - e^{-a^2} H_4(a) + H_4(0)
   \]
   As \(a \to \infty\), \(e^{-a^2} \to 0\) and \(a H_5(a) + H_4(a)\) is a polynomial of degree 6. Using L'Hôpital's rule:
   \[
   \lim_{a \to \infty} \frac{a H_5(a) + H_4(a)}{e^{a^2}} = 0
   \]
   Therefore,
   \[
   \lim_{a \to \infty} S_6(a) = H_4(0)
   \]
   From Part (1), \(H_4(x) = 16x^4 - 48x^2 + 12\), so \(H_4(0) = 12\).

The final answer is \(\boxed{12}\)