# Different definitions of e (Part 7): Connecting the two definitions

In this series of posts, I consider how two different definitions of the number $e$ are related to each other. The number $e$ is usually introduced at two different places in the mathematics curriculum:

1. Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$.
2. Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$, so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$.

These two definitions appear to be very, very different. One deals with making money. The other deals with the area under a hyperbola. Amazingly, these two definitions are related to each other. In this series of posts, I’ll discuss the connection between the two.

We begin with the second definition, which is usually considered the true definition of $e$. From this definition, I have shown in a previous post that we can derive the differentiation formulas

$\displaystyle \frac{d}{dx} (\ln x) = \frac{1}{x} \qquad$ and $\qquad \displaystyle \frac{d}{dx} \left( e^x \right) = e^x$

beginning with this definition of the number $e$.

Theorem. $\displaystyle \lim_{h \to 0} \left( 1 + h \right)^{1/h} = e$.

Proof #1. Recall the definition of a derivative

$f'(x) = \displaystyle \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

Let’s apply this to the function $f(x) = \ln x$:

$\displaystyle \frac{d}{dx} (\ln x) = \displaystyle \lim_{h \to 0} \frac{\ln(x+h) - \ln x}{h}$

$\displaystyle \frac{1}{x} = \displaystyle \lim_{h \to 0} \frac{1}{h} \ln \left( \frac{x+h}{x} \right)$

$\displaystyle \frac{1}{x} = \displaystyle \lim_{h \to 0} \frac{1}{h} \ln \left( 1 + \frac{h}{x} \right)$

$\displaystyle \frac{1}{x} = \displaystyle \lim_{h \to 0} \ln \left(1 + \frac{h}{x} \right)^{1/h}$

(I’ll note parenthetically that I’ll need the above line for a future post in this series.) At this point, let’s substitute $x = 1$:

$1 = \displaystyle \lim_{h \to 0} \ln (1 + h)^{1/h}$

Let’s now apply the exponential function to both sides:

$e^1 = \exp \left[ \displaystyle \lim_{h \to 0} \ln (1 + h)^{1/h} \right]$

Since $g(x) = e^x$ is continuous, we can interchange the function and the limit on the right-hand side:

$e = \displaystyle \lim_{h \to 0} \exp \left[ \ln (1 + h)^{1/h} \right]$

Finally, since $g(x) = e^x$ and $f(x) = \ln x$ are inverse functions, we can conclude

$e = \displaystyle \lim_{h \to 0} (1 + h)^{1/h}$.

(A second proof of this theorem, using L’Hopital’s Rule, will be presented in tomorrow’s post.)

The next theorem establishes, at last, the connection between the continuous compound interest formula and the area under the hyperbola. I’ve noted that my students feel a certain sense of accomplishment after reaching this point of the exposition.

Theorem. $\displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^{n} = e$.

Proof. Though a little bit of real analysis is necessary to make this rigorous, we can informally see why this has to be true by letting $n = 1/h$ for $h$ positive. Then the expression $(1+h)^{1/h}$ becomes $\left( 1 + \frac{1}{n} \right)^n$. Also, as $h \to 0$, then $n \to \infty$.