Show that for any positive number $\varepsilon$ we can find a positive number $\delta$ such that the statement below is satisfied.
-$6 \mathrm{x}-7$ is less than $\varepsilon$ away from 5 whenever $\mathrm{x}$ is less than $\delta$ away from 2.
A) $\delta=\frac{\varepsilon}{6}$

B) $\delta=\frac{\varepsilon}{8}$

C) $\delta=\frac{\varepsilon}{4}$

D) $\delta=\frac{\varepsilon}{7}$

Question

Show that for any positive number $\varepsilon$ we can find a positive number $\delta$ such that the statement below is satisfied.
-$6 \mathrm{x}-7$ is less than $\varepsilon$ away from 5 whenever $\mathrm{x}$ is less than $\delta$ away from 2.
A) $\delta=\frac{\varepsilon}{6}$

B) $\delta=\frac{\varepsilon}{8}$

C) $\delta=\frac{\varepsilon}{4}$

D) $\delta=\frac{\varepsilon}{7}$

Quizplus · Accepted Answer

The given condition translates to $|6x - 7 - 5| < \varepsilon$ when $|x - 2| < \delta$. Simplifying the left side gives $|6x - 12| < \varepsilon$, or $|6(x - 2)| < \varepsilon$, leading to $|x - 2| < \frac{\varepsilon}{6}$, hence $\delta = \frac{\varepsilon}{6}$.

Show That for Any Positive Number ε\varepsilonε We Can Find a Positive Number

Show That for Any Positive Number $\varepsilon$ We Can Find a Positive Number