Answer :
I suppose the integral could be
[tex]\displaystyle\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx[/tex]
In that case, since [tex]-\frac1x\to0[/tex] as [tex]x\to\infty[/tex], we know [tex]e^{-1/x}\to1[/tex]. We also have [tex]\left(e^{-1/x}\right)'=\frac{e^{-1/x}}{x^2}>0[/tex], so the integral is approach +1 from below. This tells us that, by comparison,
[tex]\displaystyle\frac{e^{-1/x}}{x^2}\le\frac1{x^2}\implies\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx\le\int_7^\infty\frac{\mathrm dx}{x^2}[/tex]
and the latter integral is convergent, so this integral must converge.
To find its value, let [tex]u=-\frac1x[/tex], so that [tex]\mathrm du=\frac{\mathrm dx}{x^2}[/tex]. Then the integral is equal to
[tex]\displaystyle\int_{-1/7}^0e^u\,\mathrm du=e^0-e^{-1/7}=1-\frac1{\sqrt[7]{e}}[/tex]