Quick Little Calculus Puzzle

fluffy, one of my friends and regular readers, got to discussing with me a couple of limit problems, particularly, ones that seemed to be solved through L’Hopital’s Rule and then ran across some that don’t call for that tool of Freshman Calculus which you maybe remember. It’s the thing about limits of zero divided by zero, or infinity divided by infinity. (It can also be applied to a couple of other “indeterminate forms”; I remember when I took this level calculus the teacher explaining there were seven such forms. Without looking them up, I think they’re \frac00, \frac{\infty}{\infty}, 0^0, \infty^{0}, 0^{\infty}, 1^{\infty}, \mbox{ and } \infty - \infty but I would not recommend trusting my memory in favor of actually studying for your test.)

Anyway, fluffy put forth two cute little puzzles that I had immediate responses for, and then started getting plagued by doubts about, so I thought I’d put them out here for people who want the recreation. They’re both about taking the limit at zero of fractions, specifically:

\lim_{x \rightarrow 0} \frac{e^x}{x^e}

\lim_{x \rightarrow 0} \frac{x^e}{e^x}

where e here is the base of the natural logarithm, that is, that number just a little high of 2.71828 that mathematicians find so interesting even though it isn’t pi.

The limit is, if you want to be exact, a subtly and carefully defined idea that took centuries of really bright work to explain. But the first really good feeling that I really got for it is to imagine a function evaluated at the points near but not exactly at the target point — in the limits here, where x equals zero — and to see, if you keep evaluating x very near zero, are the values of your expression very near something? If it does, that thing the expression gets near is probably the limit at that point.

So, yes, you can plug in values of x like 0.1 and 0.01 and 0.0001 and so on into \frac{e^x}{x^e} and \frac{x^e}{e^x} and get a feeling for what the limit probably is. Saying what it definitely is takes a little more work.