Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 31

(a) Guess the value of $\displaystyle \lim \limits_{x \to 0} \frac{x}{\sqrt{1 + 3x} - 1}$ by graphing the function $f\displaystyle (x) = \frac{x}{\sqrt{1 + 3x} - 1}$









Based on the graph, the $\lim\limits_{x \to 0} f(x)$ is approximately equal to 0.66


(b) Estimate the value of the limit by making a table of values of $f(x)$ as $x$ approaches .


$\begin{array}{|c|c|c|c|c|}
\hline
x & f(x) \\
\hline
0.01 & 0.67\\
0.02 & 0.67 \\
0.03 & 0.68 \\
0.04 & 0.69\\
\hline
\end{array} $

Based on the values from the table, the limit of the function seems to have a value of 0.67 as $x$ approaches to 0.
(c) Prove that your guess is correct by using the limit laws.



$
\begin{equation}
\begin{aligned}

& \lim \limits_{x \to 0} \frac{x}{\sqrt{1 + 3x } - 1} \cdot \frac{\sqrt{1 + 3x} + 1}{\sqrt{1 + 3x} + 1}
= \lim \limits_{x \to 0} \frac{x(\sqrt{1 + 3x} + 1)}{1 + 3x - 1}
&& \text{ Simplify the equation by multiplying both numerator and denominator by $\sqrt{1 + 3x} + 1$}\\

& \lim \limits_{x \to 0} \left( \frac{\sqrt{1 + 3x} + 1}{3} \right)
= \frac{1}{3}\lim \limits_{x \to 0} (\sqrt{1 + 3x} + 1)
&& \text{ Constant multiple law.}\\

& \lim \limits_{x \to 0} \left( \frac{\sqrt{1 + 3x} + 1}{3} \right)
= \frac{1}{3} \left( \sqrt{\lim \limits_{x \to 0} 1 + \lim \limits_{x \to 0} 3x} + \lim \limits_{x \to 0} 1 \right)
&& \text{ Sum and root law.}\\

& \lim \limits_{x \to 0} \left( \frac{\sqrt{1 + 3x} + 1}{3} \right)
= \frac{1}{3} \left( \sqrt{1 + 3 \lim \limits_{x \to 0}} + 1 \right)
&& \text{ Sum, constant multiple and special limit law.}\\

& \lim \limits_{x \to 0} \left( \frac{\sqrt{1 + 3x} + 1}{3} \right)
= \frac{1}{3} \left( \sqrt{1 + 3(0)} + 1 \right)
&& \text{ Special limit and constant law.}\\

& \fbox{$ \lim \limits_{x \to 0} \displaystyle \left( \frac{\sqrt{1 + 3x} + 1}{3} \right) = \frac{2}{3} \text{ or } 0.67$}


\end{aligned}
\end{equation}
$