Q. 69 - Calculus

Question

Prove that $\underset{x\to 2}{\lim }(7-x)=5$, with these steps:

(a) What is the $\delta -\epsilon$ statement that must be shown to prove that $\underset{x\to 2}{\lim }(7-x)=5$?

(b) Argue that $x\in (2-\delta , 2) \cup (2, 2+\delta )$ if and only if$-\delta <x-2<\delta$. Then use algebra to show that this means that $0<\left|x-2\right|<\delta$.

(c) Argue that $7-x\in (5-\epsilon , 5+\epsilon )$ if and only if $-\epsilon <2-x<\epsilon$. Then use algebra to show that this means that $\left|x-2\right|<\epsilon$.

(d) Given any particular $\epsilon >0$, what value of $\delta$ would guarantee that if $0<\left|x-2\right|<\delta$, then $\left|x-2\right|<\epsilon$? Your answer will depend on $\epsilon$.

(e) Put the previous four parts together to prove the limit statement.

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{Part} \left(\mathrm{a}\right) \mathrm{For} \mathrm{all} \mathrm{\epsilon } > 0, \mathrm{there} \mathrm{exists} \delta > 0 \mathrm{such} \mathrm{that} \mathrm{if} x \in (2 - \delta , 2) \cup (2, 2 + \delta ), \\ \mathrm{then} 7 - x \in (5 - \epsilon , 5 + \epsilon ). \\ \mathrm{Part} \left(\mathrm{b}\right) 2 -\delta < x < 2 + \delta \mathrm{and} x \ne 2 \mathrm{means} \mathrm{that} -\delta < x - 2 < \delta \mathrm{and} x \ne 2. \\ \mathrm{This} \mathrm{means} \mathrm{that} |x - 2| < \delta \mathrm{and} x - 2 \ne 0, \mathrm{and} \mathrm{thus} 0 < |x - 2| < \delta . \\ \mathrm{Part} \left(\mathrm{c}\right) 5 -\epsilon < 7 - x < 5 + \epsilon \mathrm{means} \mathrm{that} -\epsilon < 2 - x < \epsilon , \\ \mathrm{and} \mathrm{thus} \mathrm{that} |2 - x| < \epsilon , \mathrm{which} \mathrm{is} \mathrm{equivalent} \mathrm{to} |x - 2| < \epsilon . \\ \mathrm{Part} \left(\mathrm{d}\right) \mathrm{For} \mathrm{a} \mathrm{given} \epsilon > 0, \mathrm{choose} \delta = \epsilon . \\ \mathrm{Part} \left(\mathrm{e}\right) \mathrm{Given} \mathrm{any} \epsilon > 0, \mathrm{let} \delta = \epsilon . \\ \mathrm{If} x \in (2-\delta , 2) \cup (2, 2+\delta ), \\ \mathrm{then} 0 < |x -2| < \delta \mathrm{by} \mathrm{part} \left(\mathrm{b}\right). \\ \mathrm{Thus} \mathrm{by} \mathrm{part} \left(\mathrm{d}\right), |x - 2| < \epsilon , \\ \mathrm{and} \mathrm{therefore} 7 - x \in (5 - \epsilon , 5 + \epsilon ) \mathrm{by} \mathrm{part} \left(\mathrm{c}\right). \end{gathered}$$

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

$$\begin{gathered} \mathrm{For} \mathrm{all} \mathrm{\epsilon } > 0, \mathrm{there} \mathrm{exists} \delta > 0 \mathrm{such} \mathrm{that} \mathrm{if} x \in (2 - \delta , 2) \cup (2, 2 + \delta ), \\ \mathrm{then} 7 - x \in (5 - \epsilon , 5 + \epsilon ). \end{gathered}$$

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

$$\begin{gathered} \mathrm{Now}, 2 -\mathrm{\delta } < \mathrm{x} < 2 + \mathrm{\delta } \mathrm{and} \mathrm{x} \ne 2 \mathrm{means} \mathrm{that} -\mathrm{\delta } < \mathrm{x} - 2 < \mathrm{\delta } \mathrm{and} \mathrm{x} \ne 2. \\ \mathrm{This} \mathrm{means} \mathrm{that} |\mathrm{x} - 2| < \mathrm{\delta } \mathrm{and} \mathrm{x} - 2 \ne 0, \mathrm{and} \mathrm{thus} 0 < |\mathrm{x} - 2| < \mathrm{\delta }. \end{gathered}$$

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

$$\begin{gathered} \mathrm{Again}, 5 -\mathrm{\epsilon } < 7 - \mathrm{x} < 5 + \mathrm{\epsilon } \mathrm{means} \mathrm{that} -\mathrm{\epsilon } < 2 - \mathrm{x} < \mathrm{\epsilon }, \\ \mathrm{and} \mathrm{thus} \mathrm{that} |2 - \mathrm{x}| < \mathrm{\epsilon }, \mathrm{which} \mathrm{is} \mathrm{equivalent} \mathrm{to} |\mathrm{x} - 2| < \mathrm{\epsilon }. \end{gathered}$$

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

$$\begin{gathered} \mathrm{Given} \mathrm{any} \epsilon > 0, \mathrm{let} \delta = \epsilon . \\ \mathrm{If} x \in (2-\delta , 2) \cup (2, 2+\delta ), \\ \mathrm{then} 0 < |x -2| < \delta \mathrm{by} \mathrm{part} \left(\mathrm{b}\right). \\ \mathrm{Thus} \mathrm{by} \mathrm{part} \left(\mathrm{d}\right), |x - 2| < \epsilon , \\ \mathrm{and} \mathrm{therefore} 7 - x \in (5 - \epsilon , 5 + \epsilon ) \mathrm{by} \mathrm{part} \left(\mathrm{c}\right). \end{gathered}$$