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...



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



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