Wednesday, January 25, 2017

Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 19

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-3| < \delta \qquad \text{ then } \qquad \left|\frac{x}{5} - \frac{3}{5}\right| < \varepsilon\\ \end{aligned} \end{equation}



\begin{equation} \begin{aligned} & \text{But, } \\ & \phantom{x} & \left|\frac{x}{5} - \frac{3}{5}\right| = \left|\frac{1}{5} (x-3)\right| = \frac{1}{5}|x-3| \\ & \text{So, we want}\\ & \phantom{x} & \text{ if } 0 < |x-3| < \delta \qquad \text{ then } \qquad \frac{1}{5}|x-3| < \varepsilon\\ & \text{That is,} \\ & \phantom{x} & \text{ if } 0 < |x-3| < \delta \qquad \text{ then } \qquad |x-3| < 5\varepsilon\\ \end{aligned} \end{equation}


The statement suggests that we should choose \displaystyle \delta = 5 \varepsilon

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



\begin{equation} \begin{aligned} \text{if } 0 < |x-3| < \delta \text{ then, }\\ \left|\frac{x}{5} - \frac{3}{5}\right| & = \left| \frac{1}{5} (x-3) \right| = \frac{1}{5}|x-3| < \frac{\delta}{5} = \frac{5 \varepsilon}{5 } = \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}{5} - \frac{3}{5}\right| < \varepsilon\\ & \text{Therefore, by the definition of a limit}\\ & \phantom{x} \qquad \lim\limits_{x \to 3}\frac{x}{5} = \frac{3}{5} \end{aligned} \end{equation}

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