# 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.

## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there.

## 6 thoughts on “Quick Little Calculus Puzzle”

1. Substituting x=ye makes those much easier!

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1. I like the way you think.

(I might give a couple pity points if a student had no idea what to do but tossed off a gag like that.)

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2. Nearly missed this gem! I am trying to expand both nominator and denominator in Taylor’s series (and apply L’Hospital’s rule then… or is this way too sloppy?). But I haven’t figured it out yet. Am I on the right track?

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1. I don’t think that you need to expand it in a Taylor series — that’s too much work. But I’d recommend looking at the right-limit and the left-limit for both expressions first, and whether they are in the indeterminate forms L’Hopital’s Rule calls for.

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1. Yes – thanks – of course! The first one is a plain “1/0″=infinity when approached from positive x. Re Taylor’s expansion: I considered to use only terms up to certain order: If I drop anything >= O(x^2) than I end up with “1/0” again.

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1. Yeah, that you would. Unfortunately, the Taylor expansion up to a couple terms is fine for working out a feeling for how to do the real problem, but doesn’t give you what the answer is, since the higher-order powers of x are going to be biggest when x is closest to zero, which is the region of x we’re interested in.

The left-limit is much more difficult.

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