Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 20

Show that the statement $\displaystyle \lim\limits_{x \to 6} \left( \frac{x}{4} + 3 \right) = \frac{9}{2}$ is correct using the $\varepsilon$, $\delta$ definition of limit.

Based from the defintion,


$
\begin{equation}
\begin{aligned}

\phantom{x} \text{if } & 0 < |x - a| < \delta
\qquad \text{ then } \qquad
|f(x) - L| < \varepsilon\\

\phantom{x} \text{if } & 0 < |x-6| < \delta
\qquad \text{ then } \qquad
\left|\left( \frac{x}{4} + 3 \right) - \frac{9}{2}\right| < \varepsilon\\

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{But, } \\
& \phantom{x} & \left|\left( \frac{x}{4} + 3 \right)- \frac{9}{2}\right| = \left|\frac{x}{4}+3-\frac{9}{2}\right| = \left|\frac{x+12-18}{4}\right| = \left|\frac{x-6}{4} \right| = \left| \frac{1}{4}(x-6) \right|=\frac{1}{4}|x-6| \\
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
& \text{So, we want}\\
& \phantom{x} & \text{ if } 0 < |x-6| < \delta \qquad \text{ then } \qquad \frac{1}{4}|x-6| < \varepsilon\\
& \text{That is,} \\
& \phantom{x} & \text{ if } 0 < |x-6| < \delta \qquad \text{ then } \qquad |x-6| < 4\varepsilon\\
\end{aligned}
\end{equation}
$


The statement suggests that we should choose $\displaystyle \delta =4 \varepsilon$

By proving that the assumed value of $\delta$ will fit the definition...



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x-6| < \delta \text{ then, }\\
\left|\left( \frac{x}{4} + 3 \right) - \frac{9}{2}\right| & = \left| \frac{x}{4} + 3 - \frac{9}{2}\right| = \left| \frac{x+12-18}{4}\right| = \left| \frac{x-6}{4}\right| = \frac{1}{4} |x-6| < \frac{\delta}{4} = \frac{4 \varepsilon}{4} = \varepsilon
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \text{Thus, }\\
& \phantom{x} \quad\text{if } 0 < |x-6| < \delta \qquad \text{ then } \qquad \left|\left( \frac{x}{4} + 3 \right) - \frac{9}{2}\right| < \varepsilon\\
& \text{Therefore, by the definition of the limit}\\
& \phantom{x} \qquad \lim\limits_{x \to 6} \left( \frac{x}{4} + 3 \right) = \frac{9}{2}


\end{aligned}
\end{equation}
$