Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 22

Show that the statement $\displaystyle\lim\limits_{x \to -1.5} \frac{9-4x^2}{3+2x} = 6$ 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-(-1.5)| < \delta
\qquad \text{ then } \qquad
\left|\frac{9-4x^2}{3+2x} - 6\right| < \varepsilon\\

\end{aligned}
\end{equation}
$



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



$
\begin{equation}
\begin{aligned}
& \text{So, we want}\\
& \phantom{x} & \text{ if } 0 < |x+1.5| < \delta \qquad \text{ then } \qquad 2|x+1.5| < \varepsilon\\
& \text{That is,}\\
& \phantom{x} & \text{ if } 0 < |x+1.5| < \delta \qquad \text{ then } \qquad |x+1.5| < \frac{\varepsilon}{2}\\
\end{aligned}
\end{equation}
$


The statement suggests that we should choose $\displaystyle \delta = \frac{\varepsilon}{2}$

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



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x+1.5| < \delta \text{ then, }\\
& \phantom{x} & \left|
\frac{9-4x^2}{3+2x} - 6
\right|
= \left|
\frac{(3-2x)\cancel{(3+2x)}}{\cancel{(3+2x)}} -6
\right|
= \left|
3 - 2x - 6
\right|
= \left|
-2x-3
\right|
= \left|
-2 \left(x+\frac{3}{2}\right)
\right|
= 2\left| x + \frac{3}{2}\right|
<
2 \delta
=
2 \left( \frac{\varepsilon}{2} \right)
= \varepsilon
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \text{Thus, }\\
& \phantom{x} \quad\text{if } 0 < |x+1.5| < \delta \qquad \text{ then } \qquad \left|\frac{9-4x^2}{3+2x} - 6\right| < \varepsilon\\
& \text{Therefore, by the definition of a limit}\\
& \phantom{x} \qquad \lim\limits_{x \to -1.5}\frac{9-4x^2}{3+2x} = 6


\end{aligned}
\end{equation}
$