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.

logareagreen line

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.

 

Leave a comment

1 Comment

  1. Different definitions of e: Index | Mean Green Math

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: