How to explain the behaviour of $\int x^n dx$ near $n = -1$ to a layman? From both sides, this approaches infinity, but when evaluated exactly at $n = -1$, yields $\ln (x)$.
This seems similar to the behaviour of solutions to linear ODEs with characteristic polynomials (as the determinant approaches 0, a factor of $x$ appears next to the repeated root).
How can I explain this phenomenon to a layman? Obviously, I could show a proof of both, but that is probably not enough to be satisfactory. How does an infinity turn into a logarithm at the drop of a hat?
As an aside, is there a name for this behaviour?
 A: Another thought.  If $0<a<b$, $n\ne -1$, then
$$
\int_a^b x^n \;dx = \frac{b^{n+1}-a^{n+1}}{n+1}
$$
and the limit of that as $n \to -1$ is $\ln b - 
\ln a$.  Not infinity.
So, in a certain sense, your confusion comes from ignoring the constant of integration.
A: My answer is a very "soft" answer, but it's a soft question, and the point is to explain to laymen, so I'll go ahead.
As you mentioned, you could of course show them the proofs, which might convince them that it is true but wouldn't explain why.
I would simply skip the calculus involved and point out that $e$ itself can be arrived at by the formula:
$$\lim_{a\to\,0}{(1+n)^{\frac{1}{n}}}$$
And point out that by the "intuitive" expectancy, $1+0 = 1$, and $1$ raised to any power (even infinity) is still $1$, so you would think the answer to the above limit would be $1$.  Instead, it's $e$.
So I'd give them the "soft" (but nonetheless true) conclusion that in limits where you are dealing with infinity and zero in combination, $e$ is likely to show up.  And since $ln$ just means "log base $e$", this is just another form of $e$ showing up.
