Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 32

a.) Estimate the value of $\lim \limits_{x \to \infty} f(x)$ using the graph of $f(x) = \sqrt{3x^2 + 8x + 6} - \sqrt{3x^2 + 3x + 1}$.







Referring to the graph, the limit of $f(x)$ approaches $1.4$.

b.) Construct a table of values of $f(x)$ to estimate the limit


$\begin{array}{|c|c|}
\hline\\
x & f(x) \\
\hline\\
10 & 1.4535 \\
\hline\\
100 & 1.4446 \\
\hline\\
1000 & 1.4435 \\
\hline\\
10000 & 1.4434 \\
\hline\\
100000 & 1.4434\\
\hline
\end{array}
$


Referring to the table, the limit of $f(x)$ seems to have a value of $1.4434$.

c.) Determine the exact value of the limit



$
\begin{equation}
\begin{aligned}

\lim_{x \to \infty} f(x) =& \lim_{x \to \infty} \sqrt{3x^2 + 8x + 6} - \sqrt{3x^2 + 3x + 1}
\\
\\
=& \lim_{x \to \infty} \sqrt{3x^2 + 8x + 6} - \sqrt{3x^2 + 3x + 1} \cdot \frac{\sqrt{3x^2 + 8x + 6} + \sqrt{3x^2 + 3x + 1}}{\sqrt{3x^2 + 8x + 6} + \sqrt{3x^2 + 3x + 1}}
\\
\\
=& \lim_{x \to \infty} \frac{3x^2 + 8x + 6 - (3x^2 + 3x + 1)}{\sqrt{3x^2 + 8x + 6} + \sqrt{3x^2 + 3x + 1}}
\\
\\
=& \lim_{x \to \infty} \frac{\cancel{3x^2} + 8x + 6 - \cancel{3x^2} - 3x - 1}{\sqrt{3x^2 + 8x + 6} + \sqrt{3x^2 + 3x + 1}}
\\
\\
=& \lim_{x \to \infty} \frac{5x + 5}{\sqrt{3x^2 + 8x + 6} + \sqrt{3x^2 + 3x + 1}} \cdot \frac{\frac{1}{x}}{\frac{1}{\sqrt{x^2}}}
\\
\\
=& \lim_{x \to \infty} \frac{\displaystyle \frac{5 \cancel{x}}{\cancel{x}} + \frac{5}{x}}{\displaystyle \sqrt{\frac{3 \cancel{x^2}}{\cancel{x^2}}} + \frac{8x}{x^2} + \frac{6}{x^2} } + \sqrt{\frac{3\cancel{x^2}}{\cancel{x^2}} + \frac{3x}{x^2} + \frac{1}{x^2}}
\\
\\
=& \lim_{x \to \infty} \frac{\displaystyle 5 + \frac{5}{x}}{\displaystyle \sqrt{3 + \frac{8}{x} + \frac{6}{x^2}}}
\\
\\
=& \lim_{x \to \infty} \frac{\displaystyle 5 + \frac{5}{x}}{\displaystyle \sqrt{3 + \frac{8}{x}} + \frac{6}{x^2} + \sqrt{3 + \frac{3}{x} + \frac{3}{x} + \frac{1}{x^2}}}
\\
\\
=& \frac{\displaystyle \lim_{x \to \infty} \left( 5 + \frac{5}{x} \right) }{\displaystyle \lim_{x \to \infty} \left( \sqrt{3 + \frac{8}{x} + \frac{6}{x^2}} + \sqrt{3 + \frac{3}{x} + \frac{1}{x^2}} \right) }
\\
\\
=& \frac{\displaystyle 5 + \lim_{x \to \infty} \frac{5}{x}}{\displaystyle \sqrt{3 + \lim_{x \to \infty} \frac{8}{x} + \lim_{x \to \infty} \frac{6}{x^2}} + \sqrt{3 + \lim_{x \to \infty} \frac{3}{x} + \lim_{x \to \infty} \frac{1}{x^2} } }
\\
\\
=& \frac{5 + 0}{\sqrt{3 + 0 + 0} + \sqrt{3 + 0 + 0}}
\\
\\
=& \frac{5}{\sqrt{3} + \sqrt{3}}
\\
\\
=& \frac{5}{2 \sqrt{3}} \text{ or } 1.4434


\end{aligned}
\end{equation}
$