Chapter 7.4, Problem 57E

a.

To determine

To use tables to give a numerical argument that $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x=e$ support your argument graphically.

Answer to Problem 57E

x	y
1	2
10	2.5937
100	2.7048
1000	2.7169
10000	2.7181
100000	2.7183

Graph is,

  

Explanation of Solution

Given Information: The function is, $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x=e$

Calculation:

As per the given problem,

  $e\approx 2.718$

x	y
1	2
10	2.5937
100	2.7048
1000	2.7169
10000	2.7181
100000	2.7183

Graph is,

  

b.

To determine

For several different values of r, give numerical and graphical evidence that $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x=e^r$ .

Answer to Problem 57E

The annuity will run out of fund at $t=13.8629\text{years}$ .

Explanation of Solution

Given Information: The function is, $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x=e^r$

Calculation:

As per the given problem

If $r=2$ then $\underset{x\to \infty }{\lim }{\left(1+\frac{2}{x}\right)}^x=e^r\approx 7.3891$ . Input $Y_1=\underset{x\to \infty }{\lim }{\left(1+\frac{2}{x}\right)}^x$ into your calculator and then use the table to find values for the large values of x to verify numerically that the limit is $e^2$ :

$x$	$Y_1$
10	6.1917
100	7.2446
1000	7.3743
10000	7.3876
100000	7.389
1000000	7.3891
10000000	7.3891

To verify the limit graphically, input $Y_2=e^2$ and look at the graph of $Y_1$ (shown in blue) and $Y_2$ (shown in green). Since the horizontal asymptote of $Y_1$ is $Y_2$ then the limit is $e^2$ .

  

If $r=3$ then $\underset{x\to \infty }{\lim }{\left(1+\frac{3}{x}\right)}^x=e^3\approx 20.086$ . Input $Y_1=\underset{x\to \infty }{\lim }{\left(1+\frac{3}{x}\right)}^x$ into your calculator and then use the table to find values for the large values of x to verify numerically that the limit is $e^3$ :

$x$	$Y_1$
10	13.786
100	19.219
1000	19.996
10000	20.077
100000	20.085
1000000	20.086
10000000	20.086

To verify the limit graphically, input $Y_2=e^3$ and look at the graph of $Y_1$ (shown in blue) and $Y_2$ (shown in green). Since the horizontal asymptote of $Y_1$ is $Y_2$ then the limit is $e^3$ .

  

If $r=0.5$ then $\underset{x\to \infty }{\lim }{\left(1+\frac{0.5}{x}\right)}^x=e^{0.5}\approx 1.6487$ . Input $Y_1=\underset{x\to \infty }{\lim }{\left(1+\frac{0.5}{x}\right)}^x$ into your calculator and then use the table to find values for the large values of x to verify numerically that the limit is $e^{0.5}$ :

$x$	$Y_1$
10	1.6289
100	1.6467
1000	1.6485
10000	1.6487
100000	1.6487

To verify the limit graphically, input $Y_2=e^{0.5}$ and look at the graph of $Y_1$ (shown in blue) and $Y_2$ (shown in green). Since the horizontal asymptote of $Y_1$ is $Y_2$ then the limit is $e^{0.5}$ .

  

c.

To determine

Explain why compounding interest over smaller and smaller periods of time leads to the concept of interest compounded continuously.

Answer to Problem 57E

See calculation part.

Explanation of Solution

Given Information: Compounding interest over smaller and smaller periods of time leads to the concept of interest compounded continuously.

Calculation:

As per the given problem,

The compound interest formula is $A=P{\left(1+\frac{r}{n}\right)}^{nt}$ . Compounding interest over smaller and smaller periods means n gets larger and larger. From part (b), we know $\underset{x\to \infty }{\lim }{\left(1+\frac{r}{x}\right)}^n=e^r$ so as n gets larger and larger, then $A=P{\left[{\left(1+\frac{r}{n}\right)}^n\right]}^t\to P{\left(e^r\right)}^t=P{e}^{rt}$ which is formula for interest compounded continuously.