The problem is to compute the following limit involving an integral:

$\lim_{x \to \infty} \frac{\int_{0}^{x} t^2 e^{-t^2} \, dt}{x}.$

Steps to Solve:

1. Simplify the integral expression: Let $F(x) = \int_{0}^{x} t^2 e^{-t^2} \, dt$. The given limit becomes: $\lim_{x \to \infty} \frac{F(x)}{x}.$

2. Analyze $F(x)$ as $x \to \infty$: The function $t^2 e^{-t^2}$ tends to 0 as $t \to \infty$ due to the exponential term dominating. However, the integral $F(x)$ converges to a finite value since $\int_{0}^{\infty} t^2 e^{-t^2} \, dt$ is a known integral.

3. Compute the definite integral $\int_{0}^{\infty} t^2 e^{-t^2} \, dt$: Using substitution $u = t^2$ (so $du = 2t \, dt$), we rewrite: $\int_{0}^{\infty} t^2 e^{-t^2} \, dt = \frac{1}{2} \int_{0}^{\infty} u e^{-u} \, du.$ The integral $\int_{0}^{\infty} u e^{-u} \, du$ is a standard Gamma function result: $\int_{0}^{\infty} u e^{-u} \, du = \Gamma(2) = 1!.$ Thus: $\int_{0}^{\infty} t^2 e^{-t^2} \, dt = \frac{1}{2}.$

4. Limit as $x \to \infty$: As $x \to \infty$, $F(x) \to \frac{1}{2}$, and hence: $\lim_{x \to \infty} \frac{F(x)}{x} = \lim_{x \to \infty} \frac{\frac{1}{2}}{x} = 0.$

Final Answer:

$\boxed{0}$

Would you like a deeper explanation for any step? Here are related questions:

1. What is the Gamma function, and how does it relate to this problem?
2. Why does $\int_{0}^{\infty} t^2 e^{-t^2} \, dt$ converge to $\frac{1}{2}$?
3. What happens if the denominator $x$ is replaced by $x^n$ (general power)?
4. How does substitution $u = t^2$ simplify this integral?
5. Can L'Hôpital's rule be applied to this problem, and what would it yield?

Tip: When evaluating integrals of the form $\int_{0}^{\infty} x^n e^{-x^m} \, dx$, substitutions can transform them into Gamma function forms.