Consider the function $$ g(x) = \frac{1-e^{-2x^2}}{x^2} $$ for $x \neq 0$. Then I have to show that the improper integral $$ \int_0^\infty g(t) $$ is convergent.
My attempt
As $e^{-2x^2} \leq 1$ for all $x \in \mathbb{R}$ we have that $$ \frac{1-e^{-2x^2}}{x^2} \leq \frac{1}{x^2} $$ Thus as the limit $$ \lim_{b \rightarrow \infty} \int_1^b \frac{1}{x^2}dx = 1 $$ exists it follows from the comparison criteria that the improper integral is convergent. Is this approach alright? I think it is. Thanks for your help.
You didn't use that $e^{-2x^2} \le 1$ to write that inequality but rather the fact that $e^{-2x^2} \ge 0$.
However, you do need the fact that $e^{-2x^2} \le 1$ to conclude that $$0 \le \dfrac{1-e^{-2x^2}}{x^2}.$$
With that in place, you may use the comparison test.
Also, you've shown that the improper integral $\displaystyle\int_1^\infty g(x)dx$ exists. You also need to argue that the integral on $[0, 1]$ exists as well.
However, that follows simply because $g$ can be made continuous on $[0, 1]$ by appropriately defining it at $0$.