# Different definitions of e (Part 8): 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.

In yesterday’s post, I proved the following theorem, thus completing a long train of argument that began with the second definition of $e$ and ending with a critical step in the derivation of the continuous compound interest formula. Today, I present an alternate proof of the theorem using L’Hopital’s rule.

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

Proof #2. Let’s write the left-hand side as

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

Let’s take the natural logarithm of both sides:

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

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

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

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

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

The right-hand side follows the indeterminate form $0/0$, as so we may apply L’Hopital’s Rule. Taking the derivative of both the numerator and denominator, we find

$\ln L = \displaystyle \lim_{h \to 0} \frac{ \displaystyle ~ \frac{1}{1+h} ~}{1}$

$\ln L = \displaystyle \lim_{h \to 0} \frac{1}{1+h}$

$\ln L = \displaystyle \frac{1}{1+0} = 1$.

Therefore, the original limit is $L = e^1 = e$.

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