Question

15-18 Prove the statement using the \(\varepsilon \), \(\delta \) definition of a limit and illustrate with a diagram like figure 9.

17. \(\mathop {{\bf{lim}}}\limits_{x \to - 2} \left( { - {\bf{2}}x + {\bf{1}}} \right) = {\bf{5}}\)

Short answer

The given limit is true.

Step-by-step solution

Step 1:Assume the values of \(\varepsilon \) and \(\delta \)

Let \(\varepsilon > 0\) and \(\delta > 0\) such that, if \(0 < \left| {x - \left( { - 2} \right)} \right| < \delta \), then it can be represented as;

\(\begin{aligned}\left| {\left( { - 2x + 1} \right) - 5} \right| < \varepsilon \\\left| { - 2x - 4} \right| < \varepsilon \end{aligned}\)

Solve the inequality in step 1

Theinequality\(\left| { - 2x - 4} \right| < \varepsilon \)can be solved as:

\(\begin{aligned}\left| { - 2x - 4} \right| < \varepsilon \\2\left| {x - \left( { - 2} \right)} \right| < \varepsilon \\\left| {x - \left( { - 2} \right)} \right| < \frac{1}{2}\varepsilon \end{aligned}\)

Consider \(\delta = \frac{1}{2}\varepsilon \), then from the inequality \(0 < \left| {x - \left( { - 2} \right)} \right| < \delta \):

\(\left| {\left( { - 2x + 1} \right) - 5} \right| < \varepsilon \)

Thus, the value of the expressionis \(\mathop {\lim }\limits_{x \to 4} \left( { - 2x + 1} \right) = 5\).

Illustrate the limit using the diagram

Use the following steps to plot the graph of the given functions:

1. In the graphing calculator, select “STAT PLOT” and enter the equation \( - 2X + 1\) in the \({Y_1}\) tab.
2. Enter the graph button in the graphing calculator.

The figure below illustrates the limit using the diagram.

Thus, the diagram is obtained.