Chapter 3.1, Problem 1E

(a) How is the number e defined?

(b) Use a calculator to estimate the values of the limits

$\underset{h\to 0}{\lim }\frac{{2.7}^h-1}{h}\text{and}\underset{h\to 0}{\lim }\frac{{2.8}^h-1}{h}$

correct to two decimal places. What can you conclude about the value of e?

To determine

To define: The number e.

Answer to Problem 1E

The number e is $\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$.

Explanation of Solution

From the definition, e is the number such that $\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$.

To determine

To estimate: The values of the limits corrected to two decimal places.

Answer to Problem 1E

The value of the limits $\underset{h\to 0}{\lim }\frac{{2.7}^h-1}{h}$ and $\underset{h\to 0}{\lim }\frac{{2.8}^h-1}{h}$ are approximately 0.99 and 1.03.

The value of e must be between 2.7 and 2.8.

Explanation of Solution

Result used:

If the exponential function $f\left(x\right)=b^x$ is differential at 0, then it is differentiable everywhere and $f^{\prime }\left(x\right)=f^{\prime }\left(0\right)b^x,\text{ where }{f}^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\frac{b^h-1}{h}$.

Calculation:

Obtain the value of the limit $\underset{h\to 0}{\lim }\frac{{2.7}^h-1}{h}$.

h	$\frac{{2.7}^h-1}{h}$
0.1	1.04
0.01	1.00
0.001	0.99
0.0001	0.99
0.00001	0.99

Therefore, the value of limit $\underset{h\to 0}{\lim }\frac{{2.7}^h-1}{h}$ is approximately 0.99.

Obtain the value of the limit $\underset{h\to 0}{\lim }\frac{{2.8}^h-1}{h}$.

h	$\frac{{2.8}^h-1}{h}$
0.1	1.08
0.01	1.03
0.001	1.03
0.0001	1.03
0.00001	1.03

Therefore, the value of the limit $\underset{h\to 0}{\lim }\frac{{2.8}^h-1}{h}$ is approximately 1.03.

Obtain the value of e.

Consider the exponential functions ${2.7}^x\text{ and }{2.8}^x$.

Substitute 2.7 for b in $f\left(x\right)=b^x$ and apply the Result,

$f^{\prime }\left(x\right)=f^{\prime }\left(0\right){\left(2.7\right)}^x,f^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\frac{{2.7}^h-1}{h}$

Substitute 2.8 for b in $f\left(x\right)=b^x$ and apply the Result,

$f^{\prime }\left(x\right)=f^{\prime }\left(0\right){\left(2.8\right)}^x,f^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\frac{{2.8}^h-1}{h}$

Since $\underset{h\to 0}{\lim }\frac{{2.7}^h-1}{h}\approx 0.99\text{ and }\underset{h\to 0}{\lim }\frac{{2.8}^h-1}{h}\approx 1.03$, the respective derivative function is $f^{\prime }\left(x\right)=0.99{\left(2.7\right)}^x\text{ and }{f}^{\prime }\left(x\right)=1.03{\left(2.8\right)}^x$.

From the Result, the simplest differential formula exists when $f^{\prime }\left(0\right)=1$ for all possible values of b in $f\left(x\right)=b^x$.

Since $f^{\prime }{\left(0\right)}_{b=2.7}<1<f^{\prime }{\left(0\right)}_{b=2.8}$, there is a number b between 2.7 and 2.8 for which $f^{\prime }\left(0\right)=1$ and that number b is called as e.

Therefore, the number e must be between 2.7 and 2.8.