Q. 4 In this section we learned that ... [FREE SOLUTION]

Chapter 1: Q. 4 (page 151)

In this section we learned that e can be thought of as the following limit:
$\lim_{h 鈫� 0} {(1 + h)}^{\frac{1}{h}} = e .$
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
$\lim_{n 鈫� 鈭�} {(1 + \frac{1}{n})}^{n} = e .$

Short Answer

Expert verified

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

$\lim_{n 鈫� 鈭�} {(1 + \frac{1}{n})}^{n} = e \lim_{\frac{1}{h} 鈫� 鈭�} {(1 + \frac{1}{(\frac{1}{h})})}^{(\frac{1}{h})} = e \lim_{h 鈫� 0} {(1 + h)}^{\frac{1}{h}} = e$

Step by step solution

Step 1. Given information

The given limit statement is $\lim_{h 鈫� 0} {(1 + h)}^{\frac{1}{h}} = e .$

The given preceding limit statement that needs to justify is $\lim_{n 鈫� 鈭�} {(1 + \frac{1}{n})}^{n} = e .$

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

Step 2. Justification.

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

$\lim_{n 鈫� 鈭�} {(1 + \frac{1}{n})}^{n} = e \lim_{\frac{1}{h} 鈫� 鈭�} {(1 + \frac{1}{(\frac{1}{h})})}^{(\frac{1}{h})} = e \lim_{h 鈫� 0} {(1 + h)}^{\frac{1}{h}} = e$

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

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Short Answer

Step by step solution

Step 1. Given information

Step 2. Justification.

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