Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 18

Using the definition of limit of $\delta$, $\varepsilon$ prove $\lim\limits_{x \to 4} (7-3x) = -5$ and 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 - 4| < \delta
\qquad \text{ then } \qquad
|(7-3x)-(-5)| < \varepsilon\\

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{But, } \\
& \phantom{x} & |(7x-13x)-(-5)| = |7-3x+5| = |-3x+12| = |-3(x-4)| =3|x-4| \\
& \text{So, we want}\\
& \phantom{x} & \text{ if } 0 < |x-4| < \delta \qquad \text{ then } \qquad 3|x-4| < \varepsilon\\
& \text{That is,} \\
& \phantom{x} & \text{ if } 0 < |x-4| < \delta \qquad \text{ then } \qquad |x-4| < \frac{\varepsilon}{3}\\


\end{aligned}
\end{equation}
$


The statement suggests that we should choose $\displaystyle \delta = \frac{\varepsilon}{3}$

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



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x-4| < \delta \text{ then, }\\
|(7-3x)-(-5)| & = |7-3x+5| = |-3x+12| = |-3(x-4)| = 3|x-4| < 3 \delta = 3 \left(\frac{\varepsilon}{3}\right) = \varepsilon
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \text{Thus, }\\
& \phantom{x} \quad\text{if } 0 < |x-4| < \delta \qquad \text{ then } \qquad |(7x-3x)-(-5)| < \varepsilon\\
& \text{Therefore, by the definition of a limit}\\
& \phantom{x} \qquad \lim\limits_{x \to 4}(7x-3x)= -5


\end{aligned}
\end{equation}
$