Question

(a) Find a number \(\delta \) such that if \(\left| {x - {\bf{2}}} \right| < \delta \), then \(\left| {{\bf{4}}x - {\bf{8}}} \right| < \varepsilon \), where \(\varepsilon = {\bf{0}}.{\bf{1}}\).

(b) Repeat part (a) with \(\varepsilon = {\bf{0}}.{\bf{01}}\).

Short answer

a. The number is 0.025.

b. The number is 0.0025.

Step-by-step solution

Step 1:Find the number for part(a)

Obtain the number \(\delta \) by using theequation\(\left| {4x - 8} \right| < 0.1\) for \(\varepsilon = 0.1\).

\(\begin{aligned}4\left| {x - 2} \right| < 0.1\\\left| {x - 2} \right| < \frac{{0.1}}{4}\end{aligned}\)

By using the given relation \(\left| {x - 2} \right| < \delta \), the number \(\delta \) is shown below:

\(\begin{aligned}\delta < \frac{{0.1}}{4}\\ < 0.025\end{aligned}\)

So, the number\(\delta \) is 0.025.

Find the number for part (b)

Obtain the number \(\delta \) by using the equation\(\left| {4x - 8} \right| < 0.01\) for \(\varepsilon = 0.01\).

\(\begin{aligned}{c}4\left| {x - 2} \right| < 0.01\\\left| {x - 2} \right| < \frac{{0.01}}{4}\end{aligned}\)

By using the givenrelation\(\left| {x - 2} \right| < \delta \), the number \(\delta \) is shown below:

\(\begin{aligned}{c}\delta < \frac{{0.01}}{4}\\ < 0.0025\end{aligned}\)

So, the number \(\delta \) is 0.0025.