Friday, November 13, 2015

Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 20

Show that the statement lim 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-6| < \delta \qquad \text{ then } \qquad \left|\left( \frac{x}{4} + 3 \right) - \frac{9}{2}\right| < \varepsilon\\ \end{aligned} \end{equation}



\begin{equation} \begin{aligned} & \text{But, } \\ & \phantom{x} & \left|\left( \frac{x}{4} + 3 \right)- \frac{9}{2}\right| = \left|\frac{x}{4}+3-\frac{9}{2}\right| = \left|\frac{x+12-18}{4}\right| = \left|\frac{x-6}{4} \right| = \left| \frac{1}{4}(x-6) \right|=\frac{1}{4}|x-6| \\ \end{aligned} \end{equation}


\begin{equation} \begin{aligned} & \text{So, we want}\\ & \phantom{x} & \text{ if } 0 < |x-6| < \delta \qquad \text{ then } \qquad \frac{1}{4}|x-6| < \varepsilon\\ & \text{That is,} \\ & \phantom{x} & \text{ if } 0 < |x-6| < \delta \qquad \text{ then } \qquad |x-6| < 4\varepsilon\\ \end{aligned} \end{equation}


The statement suggests that we should choose \displaystyle \delta =4 \varepsilon

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



\begin{equation} \begin{aligned} \text{if } 0 < |x-6| < \delta \text{ then, }\\ \left|\left( \frac{x}{4} + 3 \right) - \frac{9}{2}\right| & = \left| \frac{x}{4} + 3 - \frac{9}{2}\right| = \left| \frac{x+12-18}{4}\right| = \left| \frac{x-6}{4}\right| = \frac{1}{4} |x-6| < \frac{\delta}{4} = \frac{4 \varepsilon}{4} = \varepsilon \end{aligned} \end{equation}



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

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