Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 16

Show that the statement $\lim\limits_{x \to -2} \displaystyle \left(\frac{1}{2}+3\right) = 2$ is correct using the
$\varepsilon$, $\delta$ definition of limit and illustrate its graph.






Based from the defintion,


$\begin{equation} \begin{aligned} \phantom{x} \text{if } & 0 < |x - a| < \delta \qquad \text{ then } \qquad |f(x) - L| < \varepsilon\\ \phantom{x} \text{if } & 0 < |x - (-2)| < \delta \qquad \text{ then } \qquad \left|\left( \frac{x}{2} + 3 \right) - 2 \right| < \varepsilon\\ \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} & \text{But, } \\ & \phantom{x} & \left|\left(\frac{x}{2}+3\right) - 2\right| = \left|\frac{x}{2}+1\right| = \left|\frac{1}{2}(x+2)\right| = \frac{1}{2}|x+2| \\ & \text{So, we want}\\ & \phantom{x} & \text{ if } 0 < |x+2| < \delta \qquad \text{ then } \qquad \frac{1}{2}|(x+2)| < \varepsilon\\ & \text{That is,} \\ & \phantom{x} & \text{ if } 0 < |x+2| < \delta \qquad \text{ then } \qquad |x+2| < 2 \varepsilon\\ \end{aligned} \end{equation}$


The statement suggests that we should choose $\displaystyle \delta = 2\varepsilon$.

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



$\begin{equation} \begin{aligned} \text{if } 0 < |x+2| < \delta \text{ then, }\\ \left|\left(\frac{x}{2}+3\right) - 2\right| & = \left|\frac{x}{2}+1\right| = \frac{1}{2}|x+2| < \frac{\delta}{2} = \frac{1}{\cancel{2}} (\cancel{2}\varepsilon) = \varepsilon \end{aligned} \end{equation}$



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