There are two different ways I've been taught to understand the meaning of $e$.
- $e$ can be defined as the total growth from continuously compounding interest in a single period. To make the growth rate the same as the dollar amount, consider I start with $\$1$. Then, $a_n$ represents the amount I have after compounding $n$ times.
\begin{align}
a_n&=1\cdot\left(1+\frac{1}{n}\right)^n\\[5pt]
e=a_{\max}&=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n
\end{align}
As we increase the number of times we compound the interest, the final amount (in the case that we start with $\$1$), approaches $e$. - All exponential functions increase at a rate proportional to their current value. For example, consider the derivative of $2^x$:
\begin{align}
\frac{d\left(2^x\right)}{dx}&=\lim_{h \to 0}\frac{2^{x+h}-2^x}{h}\\[5pt]
&=\lim_{h \to 0}\frac{2^x2^h-2^x}{h}\\[5pt]
&=\lim_{h \to 0}\frac{2^x\left(2^h-1\right)}{h}\\[5pt]
&=2^x\lim_{h \to 0}\frac{\left(2^h-1\right)}{h}
\end{align}
From this we can see that the rate at which $2^x$ changes is proportional to its current value, with a proportionality constant of $$\lim_{h \to 0}\frac{\left(2^h-1\right)}{h}.$$ For $2^x$, this constant happens to be $\ln(2)$ (why is this the case?). $e$ can be defined as the base where this proportionality constant is $1$; therefore, the rate $e^x$ grows at is exactly its current value.
My question is this: how can the same constant have two (apparently) separate definitions? Is there a way of looking at $e$ that will combine these two definitions, and make it obvious as to why they are true? Also, why is the proportionality constant of any exponential functions growth rate the natural log of its base? Is there a way to understand why this is the case intuitively?
Best Answer
Let's start with your second definition. We want to find the derivative of $f(x) = a^x$, for some $a>0.$ Using the definition of derivative, we get
\begin{align} f'(x) &= \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}\\ &= \lim_{h\to 0}\frac{a^{x+h} - a^{x}}{h}\\ &=a^{x}\lim_{h\to 0}\frac{a^{h} - 1}{h}. \end{align}
Now, as you said, one way to define $e$ is to define it such that $$\lim_{h\to 0}\frac{e^{h} - 1}{h} = 1.\tag{$*$}$$ Now, I'm going to be somewhat informal here. Suppose we have $$\frac{e^{h} - 1}{h} = 1.$$ Then, doing some manipulation: \begin{align} \frac{e^{h} - 1}{h} &= 1\\ e^{h} - 1 &= h\\ e^{h} &= h + 1\\ e &= (h+1)^{1/h} \end{align}
Now, by taking limits we have that \begin{equation*} \lim_{h\to 0}\frac{e^{h} - 1}{h} = 1 \iff e = \lim_{h\to 0}(h+1)^{1/h}. \end{equation*}
Taking the new limit, $e = \lim_{h\to 0}(h+1)^{1/h}$, and setting $n = 1/h$, we note that $h\to 0 \implies n\to \infty$, and we can change variables to get $$e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^{n},$$ which matches your first definition.
As to understanding where the natural log comes from, i.e. why we have $$\lim_{h\to 0}\frac{a^{h}-1}{h} = \ln(a),$$ this is a straightforward consequence of $(*)$: \begin{align} \lim_{h\to 0}\frac{a^{h} - 1}{h} &= \lim_{h\to 0}\frac{e^{\ln(a^{h})}-1}{h}\\ &=\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h}\\ &=\frac{\ln(a)}{\ln(a)}\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h}\\ &=\ln(a)\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h\ln(a)}.\\ \end{align} Now, set $k = h\ln(a)$, and note that because $a > 0$, we have that $h\to 0\implies k \to 0$ as well, and so we have by $(*)$ that: \begin{align} \lim_{h\to 0}\frac{a^{h} - 1}{h} &=\ln(a)\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h\ln(a)}\\ &=\ln(a)\lim_{k\to 0}\frac{e^{k} - 1}{k}\\ &=\ln(a)\cdot 1\\ &=\ln(a). \end{align}