Chapter 3.2, Problem 91E

(a)

To determine

To calculate: $f(1),f(5),f(10),f({10}^2),f({10}^4),f({10}^6)$

Answer to Problem 91E

  $\begin{array}{l}f(1)=0,f(5)=0.3,f(10)=0.1\\ f({10}^2)=0.02,f({10}^4)=0.0004,f({10}^6)=0.000006\end{array}$

Explanation of Solution

Given information: $f(x)=\frac{\ln x}{x}$

Formula used: Use $\ln a^b=b\ln a$

Calculation:

  $f(x)=\frac{\ln x}{x}$

So,

  $\begin{array}{l}f(1)=\frac{\ln 1}{1}=0\\ f(5)=\frac{\ln 5}{5}=0.3\\ f(10)=\frac{\ln 10}{10}=\frac{1}{10}=0.1\\ f({10}^2)=\frac{\ln {10}^2}{{10}^2}=\frac{2\ln 10}{100}=\frac{2}{100}=0.02\\ f({10}^4)=\frac{\ln {10}^4}{{10}^4}=\frac{4\ln 10}{10000}=\frac{4}{10000}=0.0004\\ f({10}^6)=\frac{\ln {10}^6}{{10}^6}=\frac{6\ln 10}{1000000}=\frac{6}{1000000}=0.000006\end{array}$

(b)

To determine

To find: value at which $f(x)$ approaches as $x$ increases without bound

Answer to Problem 91E

$f(x)$ approaches to 0 as $x$ increases without bound

Explanation of Solution

Given information: $f(x)=\frac{\ln x}{x}$ and values of $f(x)$ at various values of $x$

From part (a),

  $\begin{array}{l}f(1)=0,f(5)=0.3,f(10)=0.1\\ f({10}^2)=0.02,f({10}^4)=0.0004,f({10}^6)=0.000006\end{array}$

Here, $0.000006<0.0004<0.02<0.1<0.3<0$

As x increases from 1 to ${10}^6$ , $f(x)$ approaches to 0

So,

  $f(x)$ approaches to 0 as $x$ increases without bound

(c)

To determine

To graph: $f(x)=\frac{\ln x}{x}$

Explanation of Solution

Given information: $f(x)=\frac{\ln x}{x}$

Graph:

At $x=1,f(x)=\frac{\ln 1}{1}=0$

At $x=10,f(x)=\frac{\ln 10}{10}=\frac{1}{10}=0.1$

  

Interpretation:As $x$ increases without bound, $f(x)$ approaches to 0.