Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 17

Show that the statement $\lim\limits_{x \to -3} (1-4x) = 13$ is correct using the
$\varepsilon$, $\delta$ definition of limit and illustrate its graph.






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 - (-3)| < \delta
\qquad \text{ then } \qquad
|(1-4x)-13| < \varepsilon\\

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{But, } \\
& \phantom{x} & |1-4x-13| = |-4x-12| = |-4(x+3)| = 4|x+3| \\
& \text{So, we want}\\
& \phantom{x} & \text{ if } 0 < |x+3| < \delta \qquad \text{ then } \qquad 4|x+3| < \varepsilon\\
& \text{That is,} \\
& \phantom{x} & \text{ if } 0 < |x+3| < \delta \qquad \text{ then } \qquad |x+3| < \frac{\varepsilon}{4}\\


\end{aligned}
\end{equation}
$


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

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



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x+3| < \delta \text{ then, }\\
|(1-4x)-13| & = |1-4x-13| = |-4x-12| = |-4(x+3)| = 4|x+3| < 4 \delta = \cancel{4} \left(\frac{\varepsilon}{\cancel{4}}\right) = \varepsilon
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \text{Thus, }\\
& \phantom{x} \quad\text{if } 0 < |x-(-3)| < \delta \qquad \text{ then } \qquad |(1-4x)-13| < \varepsilon\\
& \text{Therefore, by the definition of a limit}\\
& \phantom{x} \qquad \lim\limits_{x \to -3} (1-4x) = 13


\end{aligned}
\end{equation}
$