Question

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

16. \(\mathop {{\bf{lim}}}\limits_{x \to 2} \left( {{\bf{2}} - {\bf{3}}x} \right) = - {\bf{4}}\)

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 \(0 < \left| {x - 2} \right| < \delta \), then it can be represented as;

\(\begin{aligned}\left| {\left( {2 - 3x} \right) - \left( { - 4} \right)} \right| < \varepsilon \\\left| {6 - 3x} \right| < \varepsilon \end{aligned}\)

Solve the inequality in step 1

Theinequality\(\left| {6 - 3x} \right| < \varepsilon \)can be solved as:

\(\begin{aligned}{r}\left| {6 - 3x} \right| < \varepsilon \\3\left| {x - 2} \right| < \varepsilon \\\left| {x - 2} \right| < \frac{1}{3}\varepsilon \end{aligned}\)

Consider \(\delta = \frac{1}{3}\varepsilon \), then from the inequality \(0 < \left| {x - 2} \right| < \delta \):

\(\left| {\left( {2 - 3x} \right) - \left( { - 4} \right)} \right| < \varepsilon \)

Thus, the value of the expressionis \(\mathop {\lim }\limits_{x \to 4} \left( {2 - 3x} \right) = - 4\).

Illustrate the limit by using the diagram

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

1. In the graphing calculator, select “STAT PLOT” and enter the equation \(2 - 3X\) in the \({Y_1}\) tab.
2. Enter the graph button in the graphing calculator.

The figure below illustrates the limit as:

Thus, the diagram is obtained.