It’s come up several times now (my and Malacandra’s posts on compound interest, and treis’s post on the antiderivative of 1/x), so it’s time to deliver on my promise of showing that (1+1/n)^n approaches e as n gets arbitrarily large. (Depending on your perspective, you may actually take this limit to be the definition of e and then prove that it satisfies d/dx e^x = e^x, but in this post, I’ll be doing the opposite, taking e to be defined by this latter property (which is, in my opinion, essentially the most natural, clean, and motivated definition for it, the one which really brings out the reasons for its special properties), and then proving it to be the limit from the compound interest problem).
First, we’ll need some preliminaries. We need to see that d/dx ln(x) = 1/x. (NB: treis showed this in his above post, but did so with (essentially) the assumption that (1+1/n)^n approaches e as n gets large; we obviously need to provide a different proof, to avoid circularity.)
Consider two variables, x and y, related by y = e^x. By the definition of e, we have that dy/dx = y.
Therefore, dx/dy = 1/y. Since x is related to y by x = ln(y), we’ve just demonstrated that d/dy ln(y) = 1/y, as we set out to.
Now, let’s consider the limit of ln(1+x)/x as x approaches 0. The top goes to ln(1+0) = 0, and the bottom obviously goes to 0 as well. This gives us the indeterminate form 0/0, so we have to apply l’Hôpital’s rule, which tells us we can replace the top and bottom in this case with their derivatives, without changing the limit. Using the chain rule and our above demonstration, we see that the derivative of ln(1+x) is 1/(1+x). And, of course, the derivative of x is 1. So, after the replacement, we get (1/(1+x))/1 = 1/(1+x). As x approaches 0, this will of course approach 1. Thus, we can conclude (via l’Hôpital) that ln(1+x)/x approaches 1 as x approaches 0.
What good is that? Well, from that limit, we can conclude (using continuity) that e^(ln(1+x)/x) approaches e^1 as x approaches 0. But e^(ln(1+x)/x) = e^(ln((1+x)^(1/x))) = (1+x)^(1/x). And so we’ve shown that (1+x)^(1/x) approaches e as x approaches 0.
This is pretty much everything we want, but I’ll go just a bit furhter. Replacing x with r/n, we see that (1+r/n)^(n/r) approaches e as r/n approaches 0. This means (1+r/n)^n approaches e^r as r/n approaches 0. Holding r constant, this is essentially the same as saying that (1+r/n)^n approaches e^r as n gets arbitrarily large. Voila, this is precisely the fact which I used above in my post on compound interest.
I think this covers almost all the special properties you’re likely to run into about e and ln. The two others noted by Chronos in post #13 (the Taylor series expansion of e^x, and Euler’s equation e^(ix) = cos(x) + isin(x) in radians, leading to the consequence that e^(ipi) = -1) follow very straightforwardly from our definition that d/dx e^x = e^x, along with the differentiation rules for sin and cos in radians, once you have some knowledge of manipulation of Taylor series.
