Improper Integral Convergence: Evaluate $\int_{-9}^{0} \frac{e^{1/x}}{x^2} \, dx$

Math Problem Statement

Evaluate the improper integral $\int_{-9}^{0} \frac{e^{1/x}}{x^2} \, dx$. If the integral does not converge, then write 'divergent'.

This problem involves evaluating the improper integral

$\int_{-9}^{0} \frac{e^{1/x}}{x^2} \, dx.$

Improper Integral Behavior:
- The integral is improper because the function $\frac{e^{1/x}}{x^2}$ has a singularity at $x = 0$ . As $x \to 0^-$ , $\frac{e^{1/x}}{x^2}$ grows extremely large. Thus, the behavior near $x = 0$ must be carefully analyzed to determine convergence.
Split the Integral: Break the integral into: $\int_{-9}^{0} \frac{e^{1/x}}{x^2} \, dx = \lim_{\epsilon \to 0^-} \int_{-9}^{\epsilon} \frac{e^{1/x}}{x^2} \, dx.$
Examine Convergence: Near $x = 0^-$ , $1/x \to -\infty$ , making $e^{1/x} \to 0^+$ . However, the $x^2$ term in the denominator grows much

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