Question

Prove the statement using the \(\varepsilon \) \(\delta \)definition of a limit and illustrate with a diagram like a Figure 15.

\(\mathop {\lim }\limits_{x \to - 2} \left( {3x + 5} \right) = - 1\)

Short answer

It is proved that\(\mathop {\lim }\limits_{x \to - 2} \left( {3x + 5} \right) = - 1\), and the graph is as follows.

Step-by-step solution

Describe the given information

It is required to prove\(\mathop {\lim }\limits_{x \to - 2} \left( {3x + 5} \right) = - 1\)by using\(\varepsilon \),\(\delta \) definition.

Prove that \(\mathop {\lim }\limits_{x \to  - 2} \left( {3x + 5} \right) =  - 1\)

Consider the limit \(\mathop {\lim }\limits_{x \to - 2} \left( {3x + 5} \right) = - 1\).

Use \(\varepsilon - \delta \) definition to prove the statement \(\mathop {\lim }\limits_{x \to - 2} \left( {3x + 5} \right) = - 1\).

If \(0 < \left| {x + 2} \right| < \delta \)then \(0 < \left| {\left( {3x + 5} \right) + 1} \right| < \varepsilon \).

Simplify in the absolute value inequality as follows.

If \(0 < \left| {x + 2} \right| < \delta \)then \(0 < \left| {3x + 6} \right| < \varepsilon \).

If \(0 < \left| {x + 2} \right| < \delta \)then \(0 < \left| {3\left( {x + 2} \right)} \right| < \varepsilon \).

Use the property \(\left| {ab} \right| = \left| a \right|\left| b \right|\) to take out \(3\).

If \(0 < \left| {x + 2} \right| < \delta \)then\(0 < 3\left| {\left( {x + 2} \right)} \right| < \varepsilon \).

Divide the inequality by \(3\)

If \(0 < \left| {x + 2} \right| < \delta \)then\(0 < \left| {\left( {x + 2} \right)} \right| < \frac{\varepsilon }{3}\).

Choose\(\delta = \frac{\varepsilon }{3}\), then

Multiply each side of the inequality by\(3\).

\(\begin{array}{c}0 < 3\left| {\left( {x + 2} \right)} \right| < \varepsilon \\0 < \left| {3\left( {x + 2} \right)} \right| < \varepsilon \\0 < \left| {3x + 6} \right| < \varepsilon \\0 < \left| {3x + 5 + 1} \right| < \varepsilon \end{array}\)

Simplify further.

\(0 < \left| {\left( {3x + 5} \right) - \left( { - 1} \right)} \right| < \varepsilon \)

Therefore, it is proved that\(\mathop {\lim }\limits_{x \to - 2} \left( {3x + 5} \right) = - 1\).

Draw the graph of the function \(f\left( x \right) = 3x + 5\) locating the distance \(\delta \) and \(\varepsilon \) limits.