Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 21

Show that the statement $\displaystyle\lim\limits_{x \to 2} \frac{x^2+x-6}{x-2} = 5$ 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-2| < \delta
\qquad \text{ then } \qquad
\left|\frac{x^2+x-6}{x-2} - 5\right| < \varepsilon\\

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{But, } \\
& \phantom{x} & \left|
\frac{x^2+x-6}{x-2} - 5
\right|
= \left|
\frac{x^2+x-6-5(x-2)}{x-2}
\right|
= \left|
\frac{x^2+x-6-5x+10}{x-2}
\right|
= \left|\frac{x^2-4x+4}{x-2}
\right|
= \left|\frac{\cancel{(x-2)}(x-2)}{\cancel{x-2}}
\right|
= |x-2|\\
\end{aligned}
\end{equation}
$



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


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

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



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x-2| < \delta \text{ then, }\\
& & \left|
\frac{x^2+x-6}{x-2} - 5
\right|
= \left|
\frac{x^2+x-6-5(x-2)}{x-2}
\right|
= \left|
\frac{x^2+x-6-5x+10}{x-2}
\right|
= \left|\frac{x^2-4x+4}{x-2}
\right|
= \left|\frac{\cancel{(x-2)}(x-2)}{\cancel{x-2}}
\right|
= |x-2| < \delta = \varepsilon
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \text{Thus, }\\
& \phantom{x} \quad\text{if } 0 < |x-3| < \delta \qquad \text{ then } \qquad \left|\frac{x^2+x-6}{x-2} - 5\right| < \varepsilon\\
& \text{Therefore, by the definition of a limit}\\
& \phantom{x} \qquad \lim\limits_{x \to 2} \frac{x^2+x-6}{x-2} = 5


\end{aligned}
\end{equation}
$