To evaluate the limit

$\lim_{h \to 0} \frac{1}{h} \int_2^{2+h} e^{-x^2} \, \mathrm{d}x,$

we can recognize that this limit represents the derivative of a function defined by an integral. Specifically, let's define a function $F(h)$ as follows:

$F(h) = \int_2^{2+h} e^{-x^2} \, \mathrm{d}x.$

Then, the given limit is

$\lim_{h \to 0} \frac{F(h)}{h}.$

This expression is the definition of the derivative of $F(h)$ at $h = 0$. In other words, we have

$\lim_{h \to 0} \frac{F(h) - F(0)}{h}.$

Since $F(0)$ is

$F(0) = \int_2^2 e^{-x^2} \, \mathrm{d}x = 0,$

the limit becomes

$\lim_{h \to 0} \frac{F(h)}{h} = F'(0).$

To find $F'(h)$, we use the Fundamental Theorem of Calculus and Leibniz's rule for differentiation under the integral sign:

$F'(h) = \frac{d}{dh} \left( \int_2^{2+h} e^{-x^2} \, \mathrm{d}x \right).$

By Leibniz's rule, this derivative is

$F'(h) = e^{-(2+h)^2} \cdot \frac{d}{dh} (2 + h) - e^{-2^2} \cdot \frac{d}{dh} (2) + \int_2^{2+h} \frac{\partial}{\partial h} (e^{-x^2}) \, \mathrm{d}x.$

Since $e^{-2^2} \cdot \frac{d}{dh} (2) = 0$ and $\frac{\partial}{\partial h} (e^{-x^2}) = 0$, the expression simplifies to

$F'(h) = e^{-(2+h)^2}.$

Evaluating this at $h = 0$, we get

$F'(0) = e^{-2^2} = e^{-4}.$

Therefore, the value of the limit is

$\boxed{e^{-4}}.$

Would you like more details or have any questions?

Here are some related questions:

1. What is the Fundamental Theorem of Calculus?
2. How does Leibniz's rule apply to differentiation under the integral sign?
3. Can you explain the concept of a derivative in the context of a limit?
4. How do you evaluate definite integrals with variable limits?
5. What is the difference between $F'(x)$ and $\frac{d}{dx} F(x)$?
6. How do you interpret $e^{-x^2}$ graphically?
7. What are some real-world applications of evaluating such limits?
8. How does this problem illustrate the use of the chain rule in calculus?

Tip: When dealing with limits involving integrals, recognizing the connection to the derivative of an integral function can simplify the problem significantly.