Chapter 2.3, Problem 25E

(a)

To determine

To estimate: The value of the function when x approaches zero by graphing the function $f\left(x\right)$.

Answer to Problem 25E

The estimated value of the function when x approaches zero $f\left(x\right)$ approaches 0.6667 or $\frac{2}{3}$.

Explanation of Solution

Given:

The graph of the function $f\left(x\right)=\frac{x}{\sqrt{1+3x}-1}$.

Draw the graph of the function $f\left(x\right)=\frac{x}{\sqrt{1+3x}-1}$ by using the graphing calculator as shown below in Figure 1.

From the graph, as $x=0$, then $f\left(x\right)$ is not defined. But x approaches 0, then $f\left(x\right)$ goes to $\frac{2}{3}$.

That is, $\underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}\approx \frac{2}{3}$.

Thus, the estimated value of $\underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}$ is $\underset{\_}{\frac{2}{3}}$.

(b)

To determine

To guess: The value of the limit by using the table of values of $f\left(x\right)$ for x close to 0.

Answer to Problem 25E

The value of the limit by using the table of values of $f\left(x\right)$ for x close to 0 is $\underset{\_}{\frac{2}{3}}$.

Explanation of Solution

Calculation:

Make the table of values of $f\left(x\right)$ for x close to 0.

x	$\sqrt{1+3x}-1$	$f\left(x\right)=\frac{x}{\sqrt{1+3x}-1}$
−0.001	$\begin{array}{l}\sqrt{1+3\left(-0.001\right)}-1\\ \approx -0.0015011267\end{array}$	$\frac{-0.001}{-0.0015011267}\approx 0.6661663$
−0.000 1	$\begin{array}{l}\sqrt{1+3\left(-0.0001\right)}-1\\ \approx -0.000150011\end{array}$	$\frac{-0.0001}{-0.000150011}\approx 0.6666167$
−0.000 01	$\begin{array}{l}\sqrt{1+3\left(-0.00001\right)}-1\\ \approx -0.00001500011\end{array}$	$\frac{-0.00001}{-0.00001500011}\approx 0.6666617$
−0.000 001	$\begin{array}{l}\sqrt{1+3\left(-0.000001\right)}-1\\ \approx -0.000001500001125\end{array}$	$\frac{-0.000001}{-0.0000015000011}\approx 0.6666662$
0.000 001	$\begin{array}{l}\sqrt{1+3\left(0.000001\right)}-1\\ \approx 0.000001499998875\end{array}$	$\frac{0.000001}{0.000001499998875}\approx 0.6666672$
0.000 01	$\begin{array}{l}\sqrt{1+3\left(0.00001\right)}-1\\ \approx 0.0000149998875\end{array}$	$\frac{0.00001}{0.0000149998875}\approx 0.6666717$
0.000 1	$\begin{array}{l}\sqrt{1+3\left(0.0001\right)}-1\\ \approx 0.00014998875\end{array}$	$\frac{0.0001}{0.00014998875}\approx 0.6667167$
0.001	$\begin{array}{l}\sqrt{1+3\left(0.001\right)}-1\\ \approx 0.001498877\end{array}$	$\frac{0.001}{0.00014998875}\approx 0.6671663$

From the table, as x gets more close to 0, the value of $f\left(x\right)$ approaches 0.66667 or $\underset{\_}{\frac{2}{3}}$.

That is, $\underset{x\to 0}{\lim }\frac{x}{\sqrt{1+3x}-1}\approx \frac{2}{3}$.

Thus, the limit appears to be $\underset{\_}{\frac{2}{3}}$.

(c)

To determine

To prove: The limit value of the function is $\frac{2}{3}$.

Explanation of Solution

Given:

The limit of the function as x approaches 0 is $f\left(x\right)=\frac{x}{\sqrt{1+3x}-1}$.

Limit Laws:

Suppose that c is a constant and the limits $\underset{x\to a}{\lim }f\left(x\right)$ and $\underset{x\to a}{\lim }g\left(x\right)$ exist. Then

Limit law 1: $\underset{x\to a}{\lim }\left[f\left(x\right)+g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }g\left(x\right)$

Limit law 2: $\underset{x\to a}{\lim }\left[f\left(x\right)-g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)-\underset{x\to a}{\lim }g\left(x\right)$

Limit law 3: $\underset{x\to a}{\lim }\left[cf\left(x\right)\right]=c\underset{x\to a}{\lim }f\left(x\right)$

Limit law 4: $\underset{x\to a}{\lim }\left[f\left(x\right)g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)\cdot \underset{x\to a}{\lim }g\left(x\right)$

Limit law 5: $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)}$ if $\underset{x\to a}{\lim }g\left(x\right)\ne 0$

Limit law 6: $\underset{x\to a}{\lim }{\left[f\left(x\right)\right]}^n={\left[\underset{x\to a}{\lim }f\left(x\right)\right]}^n$ where n is a positive integer

Limit law 7: $\underset{x\to a}{\lim }c=c$

Limit law 8: $\underset{x\to a}{\lim }x=a$

Limit law 9: $\underset{x\to a}{\lim }{x}^n=a^n$ where n is a positive integer.

Limit law 10: $\underset{x\to a}{\lim }\sqrt[n]{x}=\sqrt[n]{a}$ where n is a positive integer, if n is even, assume that $a>0$.

Limit law 11: $\underset{x\to a}{\lim }\sqrt[n]{f\left(x\right)}=\sqrt[n]{\underset{x\to a}{\lim }f\left(x\right)}$ where n is a positive integer, if n is even, assume that $\underset{x\to a}{\lim }f\left(x\right)>0$.

Note 1:

The Quotient rule is not applicable directly for the function $f\left(x\right)$ because the limit of the denominator is zero.

$\begin{array}{c}\underset{x\to 0}{\lim }\left(\sqrt{1+3x}-1\right)=\underset{x\to 0}{\lim }\left(\sqrt{1+3x}\right)-\underset{x\to 0}{\lim }1\left(\text{by limit law 2}\right)\\ =\left(\sqrt{\underset{x\to 0}{\lim }\left(1+3x\right)}\right)-\underset{x\to 0}{\lim }1\left(\text{by limit law 11}\right)\\ =\sqrt{\underset{x\to 0}{\lim }1+\underset{x\to 0}{\lim }\left(3x\right)}-1\left(\text{by limit law 1 and 7}\right)\end{array}$

$\begin{array}{c}=\sqrt{1+3\underset{x\to 0}{\lim }\left(x\right)}-1\left(\text{by limit law 3}\right)\\ =\sqrt{1+3\left(0\right)}-1\left(\text{by limit law 8}\right)\\ =1-1\\ =0\end{array}$

Note 2:

“The limit may be infinite or it may be some finite value when both numerator and denominator approach 0.”

Calculation:

By note 3, take the limit x approaches 0 but $x\ne 0$.

Simplify $f\left(x\right)$ by using elementary algebra, $f\left(x\right)=\frac{x}{\sqrt{1+3x}-1}$.

Take the conjugate of the denominator and multiply and divide of $f\left(x\right)$.

$\begin{array}{c}f\left(x\right)=\frac{x}{\sqrt{1+3x}-1}\times \frac{\sqrt{1+3x}+1}{\sqrt{1+3x}+1}\\ =\frac{x\left(\sqrt{1+3x}+1\right)}{\left(\sqrt{1+3x}-1\right)\left(\sqrt{1+3x}+1\right)}\end{array}$

Use the difference of square formula,

$\begin{array}{c}f\left(x\right)=\frac{x\left(\sqrt{1+3x}+1\right)}{\left({\left(\sqrt{1+3x}\right)}^2-{\left(1\right)}^2\right)}\\ =\frac{x\left(\sqrt{1+3x}+1\right)}{\left(1+3x-1\right)}\\ =\frac{x\left(\sqrt{1+3x}+1\right)}{3x}\end{array}$

Since the limit x approaches 0 but not equal to 0, cancel the common term $x\ne 0$ of both numerator and denominator,

$\begin{array}{c}f\left(x\right)=\frac{\left(\sqrt{1+3x}+1\right)}{3}\\ =\frac{1}{3}\left(\sqrt{1+3x}+1\right)\end{array}$

Use fact 1, $f\left(x\right)=\frac{1}{3}\left(\sqrt{1+3x}+1\right)$ and $x\ne 0$, then

$\underset{x\to 0}{\lim }\left(\frac{x}{\sqrt{1+3x}-1}\right)=\underset{x\to 0}{\lim }\left(\frac{1}{3}\left(\sqrt{1+3x}+1\right)\right)$.

Use the limit laws to obtain the limit of the function as below:

$\begin{array}{c}\underset{x\to 0}{\lim }\left(\frac{1}{3}\left(\sqrt{1+3x}+1\right)\right)=\frac{1}{3}\underset{x\to 0}{\lim }\left(\sqrt{1+3x}+1\right)\left(\text{by limit law 3}\right)\\ =\frac{1}{3}\left(\underset{x\to 0}{\lim }\left(\sqrt{1+3x}\right)+\underset{x\to 0}{\lim }1\right)\left(\text{by limit law 1}\right)\\ =\frac{1}{3}\left(\left(\sqrt{\underset{x\to 0}{\lim }\left(1+3x\right)}\right)+\underset{x\to 0}{\lim }1\right)\left(\text{by limit law 11}\right)\end{array}$

$\begin{array}{c}=\frac{1}{3}\left(\left(\sqrt{\underset{x\to 0}{\lim }1+3\underset{x\to 0}{\lim }x}\right)+\underset{x\to 0}{\lim }1\right)\left(\text{by limit law 1}\right)\\ =\frac{1}{3}\left(\left(\sqrt{1+3\left(0\right)}\right)+\left(1\right)\right)\left(\text{by limit law 7, 8}\right)\\ =\frac{1}{3}\left(1+1\right)\\ =\frac{2}{3}\end{array}$

Thus, the limit of the function is $\underset{\_}{\frac{2}{3}}$.

Hence the required proof is obtained.