Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{{{e^{ - 1/x}}}}{{{x^2}}}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr
& \int_0^\infty {\frac{{{e^{ - 1/x}}}}{{{x^2}}}dx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{{e^{ - 1/x}}}}{{{x^2}}}dx} \cr
& {\text{Where}} \cr
& \int {\underbrace {{e^{ - 1/x}}}_{{e^u}}\underbrace {\left( {\frac{1}{{{x^2}}}} \right)dx}_{du}} = {e^{ - 1/x}} + C \cr
& \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{{e^{ - 1/x}}}}{{{x^2}}}dx} = \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 1/x}}} \right]_0^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 1/b}} - {e^{ - 1/0}}} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 1/b}} - {e^{ - \infty }}} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 1/b}} - 0} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 1/b}}} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& = {e^{ - 1/\infty }} \cr
& = {e^0} \cr
& = 1 \cr} $$
