Question

(a) Estimate the value of \(\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {1 + 3x} - 1}}\)by graphing the function \(f\left( x \right) = \frac{x}{{\left( {\sqrt {1 + 3x} - 1} \right)}}\).

(b) Make a table of values of \(f\left( x \right)\) for \(x\) close to 0 and guess the value of the limit.

(c) Use the Limit Laws to prove that your guess is correct.

Short answer

1. It is observed from the graph that the value of the limit is \(\frac{2}{3}\).

2. It is observed from the table that the limit of the function to be \(\frac{2}{3}\).

3. It is proved that the value of the limit is \(\frac{2}{3}\).

Step-by-step solution

Estimate the value of \(\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {1 + 3x}  - 1}}\) by graphing the function

a)

The procedure to draw the graph of the equation by using the graphing calculator is as follows:

To estimate the value of the limit, draw the graph of the function\(f\left( x \right) = \frac{x}{{\sqrt {1 + 3x} - 1}}\)by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(X/\left( {{{\left( {1 + 3X} \right)}^{1/2}} - 1} \right)\)in the tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function \(f\left( x \right) = \frac{x}{{\sqrt {1 + 3x} - 1}}\) as shown below:

It is observed from the graph that the value of the limitis \(\frac{2}{3}\).

Thus, \(\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {1 + 3x} - 3}} \approx \frac{2}{3}\).

Guess the value of the limit by using the table of values

b)

The values of \(x\) and \(f\left( x \right)\) are listed in the table as shown below:

\(x\)	\(f\left( x \right)\)
\(\begin{array}{c} - 0.001\\ - 0.0001\\ - 0.00001\\ - 0.000001\\0.000001\\0.00001\\0.0001\\0.001\end{array}\)	\(\begin{array}{l}0.6661663\\0.6666167\\0.6666617\\0.6666662\\0.6666672\\0.6666717\\0.6667167\\0.6671663\end{array}\)

It is observed from the table that left, and right limits are the same and the value of the limit to be \(\frac{2}{3}\).

 Step 3: Use the limits laws to prove the guess

c)

Limit law states that let \(c\) be a constant and the limits \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists. Then

1. \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) + g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right)\)
2. \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right)\)
3. \(\mathop {\lim }\limits_{x \to a} \left( {cf\left( x \right)} \right) = c\mathop {\lim }\limits_{x \to a} f\left( x \right)\)
4. \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right)g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot \mathop {\lim }\limits_{x \to a} g\left( x \right)\)
5. \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,{\mathop{\rm if}\nolimits} \,\,\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).
6. \(\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^n} = {\left( {\,\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right)^n}\), where \(n\) is a positive integer
7. \(\mathop {\lim }\limits_{x \to a} \sqrt[n]{{f\left( x \right)}} = \,\sqrt[n]{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}\), where \(n\) is a positive integer

(If \(n\) is even, we assume that \(\,\mathop {\lim }\limits_{x \to a} f\left( x \right) > 0\).)

1. \(\,\mathop {\lim }\limits_{x \to a} c = c\)
2. \(\,\mathop {\lim }\limits_{x \to a} x = a\)
3. \(\,\mathop {\lim }\limits_{x \to a} {x^n} = {a^n}\), where \(n\) is a positive integer
4. \(\,\mathop {\lim }\limits_{x \to a} \sqrt[n]{x} = \sqrt[n]{a}\), where \(n\) is a positive integer

(If \(n\) is even, we assume that \(a > 0\).)

Quotient law cannot be applied since the limit of the numerator is 0.

So, rationalize the denominator and evaluate the limit as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{\sqrt {1 + 3x} - 1}} \cdot \frac{{\sqrt {1 + 3x} + 1}}{{\sqrt {1 + 3x} + 1}}} \right) &=& \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt {1 + 3x} + 1} \right)}}{{\left( {1 + 3x} \right) - 1}}\\ &=& \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt {1 + 3x} + 1} \right)}}{{3x}}\\ &=& \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 3x} + 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ &=& \frac{1}{3}\left( {\sqrt {\mathop {\lim }\limits_{x \to 0} \left( {1 + 3x} \right)} + \mathop {\lim }\limits_{x \to 0} 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}\)

Solve further,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{\sqrt {1 + 3x} - 1}} \cdot \frac{{\sqrt {1 + 3x} + 1}}{{\sqrt {1 + 3x} + 1}}} \right) &=& \frac{1}{3}\left( {\sqrt {\mathop {\lim }\limits_{x \to 0} \left( {1 + 3\mathop {\lim }\limits_{x \to 0} x} \right)} + \mathop {\lim }\limits_{x \to 0} 1} \right)\,\,\,\,\,\,\\ &=& \frac{1}{3}\left( {\sqrt {\left( {1 + 3 \cdot 0} \right)} + 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ &=& \frac{1}{3}\left( {1 + 1} \right)\\ &=& \frac{2}{3}\end{array}\)

Thus, it is proved that the value of the limit is \(\frac{2}{3}\).