Question

(a) Use a graph of\(f\left( x \right) = \sqrt {3{x^2} + 8x + 6} - \sqrt {3{x^2} + 3x + 1} \)to estimate the value of\(\mathop {\lim }\limits_{x \to \infty } f\left( x \right)\)to one decimal place.

(b)Use a table of values of\(f\left( x \right)\)to estimate the limit to four decimal places.

(c) Find the exact value of the limit.

Short answer

a) The graph that the value of the limit is \(\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1.4\).

b) It is observed from the table that the value of the limit to be 1.4434.

c) The exact value of the limit is 1.443376.

Step-by-step solution

Estimate the value of \(\mathop {\lim }\limits_{x \to \infty } f\left( x \right)\) 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\(f\left( x \right) = \sqrt {3{x^2} + 8x + 6} - \sqrt {3{x^2} + 3x + 1} \)by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\sqrt {3{X^2} + 8X + 6} - \sqrt {3{X^2} + 3X + 1} \)in the\({Y_1}\)tab.

2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function \(f\left( x \right) = \sqrt {3{x^2} + 8x + 6} - \sqrt {3{x^2} + 3x + 1} \) as shown below:

It is observed from the graph of the function \(f\left( x \right) = \sqrt {3{x^2} + 8x + 6} - \sqrt {3{x^2} + 3x + 1} \) that the value of the \(\mathop {\lim }\limits_{x \to \infty } f\left( x \right)\) (to one decimal) to be \(1.4\).

Thus, the value of the limit is \(\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1.4\).

Estimate 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:

\(x\)	\(f\left( x \right)\)
\(\begin{aligned}{l}10,000\\100,000\\1,000,000\end{aligned}\)	\[\begin{aligned}{l}1.44339\\1.44338\\1.44338\end{aligned}\]

It is observed from the table that the value of the limit (to four decimal places) is to be \(1.4434\).

Prove that your guess is correct 

c)

Multiply numerator and denominator by the conjugate radical and evaluate the limit as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to \infty } f\left( x \right) &= \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {\sqrt {3{x^2} + 8x + 6} - \sqrt {3{x^2} + 3x + 1} } \right)\left( {\sqrt {3{x^2} + 8x + 6} + \sqrt {3{x^2} + 3x + 1} } \right)}}{{\sqrt {3{x^2} + 8x + 6} + \sqrt {3{x^2} + 3x + 1} }}\\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {3{x^2} + 8x + 6} \right) - \left( {3{x^2} + 3x + 1} \right)}}{{\sqrt {3{x^2} + 8x + 6} + \sqrt {3{x^2} + 3x + 1} }}\\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {5x + 5} \right)\left( {\frac{1}{x}} \right)}}{{\sqrt {3{x^2} + 8x + 6} + \sqrt {3{x^2} + 3x + 1} \left( {\frac{1}{x}} \right)}}\\ &= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {{x^2} + x + 1} \right) - {x^2}}}{{\sqrt {{x^2} + x + 1} - x}}\end{aligned}\)

Solve further,

\(\begin{aligned}\mathop {\lim }\limits_{x \to \infty } f\left( x \right) &= \mathop {\lim }\limits_{x \to \infty } \frac{{5 + \frac{5}{x}}}{{\sqrt {3 + \frac{8}{x} + \frac{6}{{{x^2}}}} + \sqrt {3 + \frac{3}{x} + \frac{1}{{{x^2}}}} }}\\ &= \frac{5}{{\sqrt 3 + \sqrt 3 }}\\ &= \frac{5}{{2\sqrt 3 }}\\ &= \frac{{5\sqrt 3 }}{6}\\ \approx 1.443376\end{aligned}\)

Thus, the exact value of the limit is \(1.443376\).