The integral provided is:

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

This is an improper integral due to the limits of integration (specifically at $x = 0$ where the function $\frac{e^{-1/x}}{x^2}$ is undefined). To evaluate this, we consider the integral as a limit:

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

Analysis

• Near $x = 0^-$, the term $e^{-1/x}$ grows extremely large because $-1/x$ approaches $-\infty$, and $\frac{1}{x^2}$ diverges as well.
• This combination indicates that the integral might diverge.

To confirm, we approximate or analyze the behavior of the integrand near $x = 0^-$. The integrand dominates due to $\frac{1}{x^2}$, which diverges as $x \to 0^-$.

Conclusion

The improper integral diverges due to the divergence at $x = 0$. Thus, the result is:

$\text{Divergent.}$

Let me know if you'd like a more detailed breakdown of the divergence or related concepts.

Here are 5 follow-up questions to expand understanding:

1. What does it mean for an improper integral to converge or diverge?
2. How does the behavior of $e^{-1/x}$ affect the convergence near $x = 0$?
3. Are there any techniques to transform divergent integrals into convergent ones?
4. What role does the limit process play in evaluating improper integrals?
5. Can we visualize this integral graphically to understand divergence?

Tip: Always check the behavior of both the integrand and its bounds when working with improper integrals.