Q. 4 - Calculus

Question

In this section we learned that e can be thought of as the following limit:

$\underset{h\to 0}{\lim }{\left(1+h\right)}^{\frac{1}{h}}=e.$

In the following exercise, you will investigate the convergence of this limit and also get a preview of the Taylor series, which we will see in Chapter 8.

Use the substitution $n=\frac{1}{h}$to show that the preceding limit statement is equivalent to the limit statement

$\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=e.$

Step-by-Step Solution

Answer

The preceding limit statement is equivalent to the limit statement as follows.

$$\begin{gathered} \underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=e \\ \underset{\frac{1}{h}\to \infty }{\lim }{\left(1+\frac{1}{\left(\frac{1}{h}\right)}\right)}^{\left(\frac{1}{h}\right)}=e \\ \underset{h\to 0}{\lim }{\left(1+h\right)}^{\frac{1}{h}}=e \end{gathered}$$

Step 1. Given information

The given limit statement is $\underset{h\to 0}{\lim }{\left(1+h\right)}^{\frac{1}{h}}=e.$

The given preceding limit statement that needs to justify is $\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=e.$

The given equation is $n=\frac{1}{h}.$

Step 2. Justification.

Substitute $n=\frac{1}{h}$in preceding limit statement.

$$\begin{gathered} \underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=e \\ \underset{\frac{1}{h}\to \infty }{\lim }{\left(1+\frac{1}{\left(\frac{1}{h}\right)}\right)}^{\left(\frac{1}{h}\right)}=e \\ \underset{h\to 0}{\lim }{\left(1+h\right)}^{\frac{1}{h}}=e \end{gathered}$$

So the preceding limit statement is equivalent to the limit statement.