A contradiction I found with integrals and limits of an exponential function "a" is a constant.
$$\lim_{a\to 0^+} \int e^{ax} \, dx = \lim_{a\to 0^+} \frac{e^{ax}}{a}=\frac{e^{0^+\times x}}{0^+}=\frac{1}{0^+}=\infty$$
but doing it in a different order:
$$\lim_{a\to 0^+} \int e^{ax} \, dx =\int e^{0^+\times x} \, dx=\int e^{0} \, dx=\int 1 \, dx=x$$
so which one is it? why did I get this contradiction? Thanks.

Integrate your function from $0$ to $x$ like this:
$$\lim_{a\to 0^+}\int_0^x e^{at} \, dt = \lim_{a\to 0^+} \bigg(\frac{1}{a} e^{ax} - \frac{1}{a}\bigg).$$
You can see how this plays out from here. I prefer definite integrals with a variable upper bound because of issues like this amongst others. I have personally become, pardon the word play, very anti antiderivative over time. (Does that mean I'm derivative?)
