How would I go about finding whether this integral diverges or converges using the direct comparison test ?
$$\int_{0}^{\ln2}\frac{e^{-1/x}}{x^2}dx$$
Is it true to say that $\frac{1}{x^2}\leq\frac{e^{-1/x}}{x^2}$? Bear in mind that I can't use a graphics calculator to see if this is true. $e^{-1/x}$ is the trouble maker for me to understand the inequality.
I'm not sure why you need to use the direct comparison test, since you can compute the integral explicitly using a $u$-substitution. Set $u = -\frac{1}{x}$, then $du = \frac{1}{x^2} dx$, so the integral is $$ \int_{x=0}^{x=\ln 2} e^u du = e^u \bigg|_{x=0}^{x=\ln x} = e^{-1/x} \bigg|_{0}^{\ln 2} $$ Now it's just a question of computing the limit $\lim_{x \to 0^+} e^{-1/x}$ and whether that is finite. As it turns out, the limit is zero.