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Let's define $e$ as the number $a$ such that $\frac {d}{dx} a^x = a^x$. I'm trying to prove that this $a$ has to be $e$. I don't see any way of proceeding from here except by the limit definition (I'm not assuming I know what the $\ln$ function is, or else there'd be a much easier definition of $e$ to be had).

$$\frac {d}{dx} a^x=\lim_{h\to 0} \frac {a^{x+h}-a^x}{h}=\lim_{h\to 0} a^x\frac{a^h-1}{h}=a^x\left(\lim_{h\to 0} \frac{a^h-1}{h}\right)$$

So clearly $e$ must be the number that makes that limit on the far right equal to $1$. I'm not sure how to evaluate this. Using L'Hopital's rule, I just get $\lim_{h\to 0} \frac {d}{dh} a^h$, which doesn't particularly help.

So my question:

How can I prove that there is a unique number such that $\lim_{h\to 0} \frac {a^h-1}{h}=1$?

user195508
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    What definition of the function $x \mapsto a^x$ are you assuming? – Rob Arthan Nov 26 '14 at 00:49
  • I was thinking the function $x \mapsto a^x$ with $x$ real and $a$ nonnegative (though as Steve Kass notes, I really should have only considered positive $a$), but a proof involving positive $a$ & complex $x$ would be cool, too. I just haven't gotten to complex analysis, so I might not understand it. – user195508 Nov 26 '14 at 02:43
  • how are you planning to define $a^x$ for arbitrary real $a$ and $x$ without using the exponential function? – Rob Arthan Nov 26 '14 at 22:50

2 Answers2

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Unfortunately, as you note, the limit $$\lim_{h\rightarrow0}\frac{a^h-1}h$$ is not easily worked with (since it equals $\log(a)$). We can, however, gleam two bits of information from it. Firstly, if we define the function $$f(a)=\lim_{h\rightarrow 0}\frac{a^h-1}h$$ we can show that it is monotonic increasing; in particular, notice that if $b>a$ then, from the continuity of $x\mapsto a^x$ we can find some $k>1$ such that $a^{k}=b$. Thus, we could write $$f(b)=\lim_{h\rightarrow 0}\frac{a^{kh}-1}{h}$$ or if we substitute $h'=kh$, we get $$f(b)=\lim_{h'\rightarrow 0}k\frac{a^{h'}-1}{h'}=kf(a)>f(a).$$ Of course, when you substitute things in, you quickly find that $f=\log$ and $k=\log_a(b)$, but we never used these facts and are, indeed, proving all this from scratch.

We could also derive a formula for $e$ knowing that $$\lim_{h\rightarrow 0}\frac{e^h-1}h=1$$ In particular, choose $h=\frac{1}n$ for integer $n$ (which is justified since $\frac{e^h-1}h$ is continuous for $h\neq 0$. Then we have $$\lim_{n\rightarrow\infty}n(\sqrt[n]e-1)=1$$ Now, for a fixed $n$ define $e_n$ to be the value such that the above equality (ignoring the limit) holds. That is $$n(\sqrt[n]{e_n}-1)=1$$ and solving for $e_n$ gives $$e_n=\left(1+\frac{1}{n}\right)^n.$$ It should be fairly clear that $e=\lim_{n\rightarrow\infty}e_n$ and this fact could be made formal without too much effort - but intuitively, it is clear, since $e_n$ is the solution for each case approaching the limit, the limit thereof is the solution for the limiting case.

Milo Brandt
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  • This argument assumes that the definition of $f(a)$ makes sense, i.e., that $\lim_{h \rightarrow 0} (a^h - 1)/h$ exists in the first place. – KCd Nov 26 '14 at 02:01
  • The question assumes this too; it's not too hard to prove though. To sketch a proof: it is clear, from the AM-GM inequality, that $x\mapsto a^x$ is convex. This implies differentiable almost everywhere. However, since for any pair of points $p_1,p_2$ on the curve $(x,a^x)$, there is an affine transformation taking $p_1$ to $p_2$ and preserving the curve, if the function is differentiable anywhere, it is differentiable everywhere. It is thus differentiable everywhere. – Milo Brandt Nov 26 '14 at 02:13
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First, note that $\frac {d}{dx} 0^x = 0^x$, so you really want to know that there is a unique positive real number $a$ for which $\frac {d}{dx} a^x = a^x$.

Suppose that $\frac{d}{dx}a^x=a^x$ and $\frac{d}{dx}b^x=b^x$, with $a$ and $b$ both positive. Then by the quotient rule, $$\frac{d}{dx}\left(\frac{a}{b}\right)^x=\frac{d}{dx}\frac{a^x}{b^x}=\frac{b^xa^x-a^xb^x}{b^{2x}}=0.$$ Then $\left(\frac{a}{b}\right)^x$ is a constant, which implies that $\frac{a}{b}$ is $1$ or $0$. Since $a>0$, $\frac{a}{b}$ must be $1$, so $a=b$.

Steve Kass
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