1/a approaches infinity I know very little calculus and I'm trying to understand this video from a MOOC I'm doing. I am trying to understand when at about 1:46 he says that $a$ doesn't approach infinity, but $1/a$ does. I though it was the exact opposite, am I taking something out of context? Please keep as little calculus in the answer as possible. Maybe one or two derivatives, integrals, or limits, but very little.
Thank you.
 A: As $a$ decreases to $0$, then $1/a$ increases to $\infty$.
Imagine $a$ is some tiny tiny microscopic number, and you ask how many times $a$ goes into $1$.  It's a very large number.  And it can be made as large as you want by making $a$ small enough.
A: When $a$ is close to $0$ the expression $e^{-ax}$ is close to $1$
which gives the intuition that 
$$
\int_{0}^{\infty}e^{-ax}\approx\int_{0}^{\infty}1\underbrace{=}_{\int_{a}^{b}c=c(b-a)}\infty
$$
and from the expressions $\frac{1}{a},1,a$ only $\frac{1}{a}$ is
large when $a$ is small.
For example if $a=0.001$ then $\frac{1}{a}=1000$.
On a side note - I think that you should be more comfortable with
limits when studying about improper integrals, maybe consider reviewing
this material 
A: The video isn't very clear because of the speaker's language and use of pronouns. What he means is that we cannot have $$ \int_{0}^{\infty} e^{-ax} \, dx = a $$ because $$ \int_{0}^{\infty} e^{-ax} \, dx \rightarrow \infty \,\, \text{as} \,\, a \rightarrow 0 $$ but, clearly, $a$ does not go to $\infty$ as $a \rightarrow 0.$ On the other hand, $\frac{1}{a} \rightarrow \infty$ as $a \rightarrow 0$ so that $\frac{1}{a}$ is a better candidate answer for the value of the integral.
