The given problem requires the evaluation of the improper integral:

$\int_{5}^{0} \frac{e^{-1/x}}{x^2} \, dx$

To solve this:

1. Change the limits: Rewriting the integral with correct orientation, this becomes: $\int_{0}^{5} \frac{e^{-1/x}}{x^2} \, dx$

   However, the integral has an improper point at $x = 0$, so we consider the limit: $\lim_{\epsilon \to 0^+} \int_{\epsilon}^{5} \frac{e^{-1/x}}{x^2} \, dx$

2. Analyze the behavior near $x = 0$: The term $\frac{e^{-1/x}}{x^2}$ has $e^{-1/x}$, which approaches $0$ as $x \to 0^+$. Therefore, the integrand diminishes rapidly near $x = 0$.

3. Substitute for integration: Use the substitution $u = -\frac{1}{x}$, implying that $du = \frac{1}{x^2} \, dx$ and when $x = 0$, $u \to -\infty$, and when $x = 5$, $u = -\frac{1}{5}$. The integral becomes: $\int_{\epsilon}^{5} \frac{e^{-1/x}}{x^2} \, dx = \int_{-\infty}^{-\frac{1}{5}} e^{u} \, du$

4. Evaluate the integral: The integral of $e^{u}$ is simply $e^{u}$, so: $\int_{-\infty}^{-\frac{1}{5}} e^{u} \, du = \left[e^{u}\right]_{-\infty}^{-\frac{1}{5}} = e^{-1/5} - e^{-\infty}$

   Since $e^{-\infty} = 0$, this simplifies to: $e^{-1/5}$

5. Final Answer: $\boxed{e^{-1/5}}$

Let me know if you'd like a detailed explanation of any step or have any further questions.

Here are five related questions for practice:

1. What is the significance of improper integrals in calculus?
2. How does the substitution $u = -1/x$ simplify this integral?
3. What happens if the integrand diverges instead of converging?
4. Can you evaluate $\int_{0}^{1} \frac{e^{-1/x}}{x^2} \, dx$ similarly?
5. How would this change if the limits of integration were $[-5, 0]$?

Tip: When dealing with improper integrals, always check the convergence of the integrand near problematic points (like $0$, $\infty$, or undefined points).