Chapter 13.2, Problem 27E

a.

To determine

To find: the value of limit by using graphing calculator.

Answer to Problem 27E

The value of limit is $0.667$ .

Explanation of Solution

Given information:

The limit is $\underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}$ .

Calculation: Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $\frac{x}{\sqrt{1+3x}-1}$ which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as:

  

From the above graph it can be observed that the value of the limit is $0.667$ .

b.

To determine

To find: the value of limit by making a table.

Answer to Problem 27E

The value of limit is $0.667$ .

Explanation of Solution

Given information:

The limit is $\underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}$ .

Calculation: to find the value of limit, substitute the values of x and find the value of function,

$x$	$f\left(x\right)$
$0.1$	$0.71339$
$0.01$	$0.6716$
$0.001$	$0.6671$

By the above table the value of limit should be $0.667$ .

c.

To determine

To find: the value of limit by using limit laws.

Answer to Problem 27E

The value of limit is $\frac{2}{3}$ .

Explanation of Solution

Given information:

The limit is $\underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}$ .

Calculation: to find the value of limit, first multiply the conjugate of the denominator and then simplify,

  $\begin{array}{l}\underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}=\underset{x\to 0}{\lim }\frac{x\left(\sqrt{1+3x}+1\right)}{\left(\sqrt{1+3x}-1\right)\left(\sqrt{1+3x}+1\right)}\text{ [multiply by conjugate]}\\ \underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}=\underset{x\to 0}{\lim }\frac{x\left(\sqrt{1+3x}+1\right)}{1+3x-1}\text{ [multiply]}\\ \underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}=\underset{x\to 0}{\lim }\frac{\sqrt{1+3x}+1}{3}\text{ [simplify]}\\ \underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}=\frac{2}{3}\end{array}$

Thus, the value of limit is $\frac{2}{3}$ .