Q. 68 - Calculus

Question

$$\begin{gathered} \mathrm{Prove} \mathrm{that} \underset{\mathrm{x}\to 1}{\lim 3}\mathrm{x}=3, \mathrm{with} \mathrm{these} \mathrm{steps}: \\ \left(\mathrm{a}\right) \mathrm{What} \mathrm{is} \mathrm{the} \delta –\epsilon \mathrm{statement} \mathrm{that} \mathrm{must} \mathrm{be} \mathrm{shown} \mathrm{to} \mathrm{prove} \mathrm{that} \underset{\mathrm{x}\to 1}{\lim 3}\mathrm{x}=3? \\ \left(\mathrm{b}\right) \mathrm{Argue} \mathrm{that} x \in (1 - \delta , 1) \cup (1, 1 + \delta ) \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} -\delta < x - 1 < \delta , \\ with x \ne 1. \mathrm{Then} \mathrm{use} \mathrm{algebra} \mathrm{to} \mathrm{show} \mathrm{that} \mathrm{this} \mathrm{means} \mathrm{that} 0 < |x - 1| < \delta . \\ \left(\mathrm{c}\right) \mathrm{Argue} \mathrm{that} 3x \in (3 - \epsilon , 3 + \epsilon ) \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} -\epsilon < 3(x - 1) < \epsilon . \\ \mathrm{Then} \mathrm{use} \mathrm{algebra} \mathrm{to} \mathrm{show} \mathrm{that} \mathrm{this} \mathrm{means} \mathrm{that} 3|x - 1| < \epsilon . \\ \left(\mathrm{d}\right) \mathrm{Given} \mathrm{any} \mathrm{particular} \mathrm{\epsilon } > 0, \mathrm{what} \mathrm{value} \mathrm{of} \mathrm{\delta } \mathrm{would} \mathrm{guarantee} \mathrm{that} \\ \mathrm{if} 0 < |x - 1| < \delta , \mathrm{then} 3|x - 1| < \epsilon ? \mathrm{Your} \mathrm{answer} \mathrm{will} \mathrm{depend} \mathrm{on} \epsilon . \\ \left(\mathrm{e}\right) \mathrm{Put} \mathrm{the} \mathrm{previous} \mathrm{four} \mathrm{parts} \mathrm{together} \mathrm{to} \mathrm{prove} \mathrm{the} \mathrm{limit} \mathrm{statement}. \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{Part} \left(\mathrm{a}\right) \mathrm{For} \mathrm{all} \mathrm{\epsilon } > 0, \mathrm{there} \mathrm{exists} \delta > 0 \mathrm{such} \mathrm{that} \mathrm{if} x \in (1 - \delta , 1) \cup (1, 1 + \delta ), \\ \mathrm{then} 3x \in (3 - \epsilon , 3 + \epsilon ). \\ \mathrm{Part} \left(\mathrm{b}\right) 1 -\delta < x < 1 + \delta \mathrm{and} x \ne 1 \mathrm{means} \mathrm{that} -1 < x - 1 < 1 \mathrm{and} x \ne 1. \\ \mathrm{This} \mathrm{means} \mathrm{that} |x - 1| < \delta \mathrm{and} x - 1 \ne 0, \mathrm{and} \mathrm{thus} 0 < |x - 1| < \delta . \\ \mathrm{Part} \left(\mathrm{c}\right) 3 -\epsilon < 3x < 3 + \epsilon \mathrm{means} \mathrm{that} -\epsilon < 3x-3 < \epsilon , \\ \mathrm{and} \mathrm{thus} \mathrm{that} |3x-3| < \epsilon , \mathrm{which} \mathrm{is} \mathrm{equivalent} \mathrm{to} 3|x - 1| < \epsilon . \\ \mathrm{Part} \left(\mathrm{d}\right) \mathrm{For} \mathrm{a} \mathrm{given} \epsilon > 0, \mathrm{choose} \delta = \epsilon . \\ \mathrm{Part} \left(\mathrm{e}\right) \mathrm{Given} \mathrm{any} \epsilon > 0, \mathrm{let} \delta = \epsilon . \\ \mathrm{If} x \in (1-\delta , 1) \cup (1, 1+\delta ), \\ \mathrm{then} 0 < |x -1| < \delta \mathrm{by} \mathrm{part} \left(\mathrm{b}\right). \\ \mathrm{Thus} \mathrm{by} \mathrm{part} \left(\mathrm{d}\right), 3|x - 1| < \epsilon , \\ \mathrm{and} \mathrm{therefore} 3x \in (3 - \epsilon , 3 + \epsilon ) \mathrm{by} \mathrm{part} \left(\mathrm{c}\right). \end{gathered}$$

Part (a) Step 1. The &#948; - &#949; statement

$$\begin{gathered} \mathrm{For} \mathrm{all} \mathrm{\epsilon } > 0, \mathrm{there} \mathrm{exists} \delta > 0 \mathrm{such} \mathrm{that} \mathrm{if} x \in (1 - \delta , 1) \cup (1, 1 + \delta ), \\ \mathrm{then} 3x \in (3 - \epsilon , 3 + \epsilon ). \end{gathered}$$

Part (b) Step 1. Argument regarding &#948;

$$\begin{gathered} \mathrm{Now}, 1 -\delta < x < 1 + \delta \mathrm{and} x \ne 1 \mathrm{means} \mathrm{that} -1 < x - 1 < 1 \mathrm{and} x \ne 1. \\ \mathrm{This} \mathrm{means} \mathrm{that} |x - 1| < \delta \mathrm{and} x - 1 \ne 0, \mathrm{and} \mathrm{thus} 0 < |x - 1| < \delta . \end{gathered}$$

Part (c) Step 1. Argument regarding &#949;

$$\begin{gathered} \mathrm{Again}, 3 -\epsilon < 3x < 3 + \epsilon \mathrm{means} \mathrm{that} -\epsilon < 3x-3 < \epsilon , \\ \mathrm{and} \mathrm{thus} \mathrm{that} |3x-3| < \epsilon , \mathrm{which} \mathrm{is} \mathrm{equivalent} \mathrm{to} 3|x - 1| < \epsilon . \end{gathered}$$

Part (d) Step 1. Relationship between &#948; - &#949;

$\mathrm{For} \mathrm{a} \mathrm{given} \epsilon > 0, \mathrm{choose} \delta = \epsilon .$

Part (e) Step 1. Proof of the limit statement

$$\begin{gathered} \mathrm{Given} \mathrm{any} \epsilon > 0, \mathrm{let} \delta = \epsilon . \\ \mathrm{If} x \in (1-\delta , 1) \cup (1, 1+\delta ), \\ \mathrm{then} 0 < |x -1| < \delta \mathrm{by} \mathrm{part} \left(\mathrm{b}\right). \\ \mathrm{Thus} \mathrm{by} \mathrm{part} \left(\mathrm{d}\right), 3|x - 1| < \epsilon , \\ \mathrm{and} \mathrm{therefore} 3x \in (3 - \epsilon , 3 + \epsilon ) \mathrm{by} \mathrm{part} \left(\mathrm{c}\right). \end{gathered}$$