Question

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

\(\mathop {\lim }\limits_{x \to 4} \left( {2x - 5} \right) = 3\)

Short answer

It is proved that\(\mathop {\lim }\limits_{x \to 4} \left( {2x - 5} \right) = 3\), and the graph is as follows.

Step-by-step solution

Describe the given information 

It is required to prove\(\mathop {\lim }\limits_{x \to 4} \left( {2x - 5} \right) = 3\)by using\(\varepsilon \),\(\delta \)definition.

Prove that \(\mathop {\lim }\limits_{x \to 4} \left( {2x - 5} \right) = 3\)

Consider the limit\(\mathop {\lim }\limits_{x \to 4} \left( {2x - 5} \right) = 3\).

Use \(\varepsilon - \delta \) definition to prove the statement\(\mathop {\lim }\limits_{x \to 4} \left( {2x - 5} \right) = 3\).

If \(0 < \left| {x - 4} \right| < \delta \)then\(0 < \left| {\left( {2x - 5} \right) - 3} \right| < \varepsilon \).

Simplify in the absolute value on the right and factor out a\(2\), so the above statement becomes,

If \(0 < \left| {x - 4} \right| < \delta \)then\(0 < \left| {2x - 8} \right| < \varepsilon \).

If \(0 < \left| {x - 4} \right| < \delta \)then\(0 < \left| {2\left( {x - 4} \right)} \right| < \varepsilon \).

Use the property \(\left| {ab} \right| = \left| a \right|\left| b \right|\) to take out\(2\).

If \(0 < \left| {x - 4} \right| < \delta \)then\(0 < 2\left| {x - 4} \right| < \varepsilon \).

Divide the inequality by\(2\).

If \(0 < \left| {x - 4} \right| < \delta \)then\(0 < \left| {x - 4} \right| < \frac{\varepsilon }{2}\).

Choose\(\delta = \frac{\varepsilon }{2}\), and then multiply each side of the inequality by\(2\).

\(\begin{array}{c}0 < 2\left| {x - 4} \right| < \varepsilon \\0 < \left| {2\left( {x - 4} \right)} \right| < \varepsilon \\0 < \left| {2x - 8} \right| < \varepsilon \\0 < \left| {2x - 5 - 3} \right| < \varepsilon \end{array}\)

Simplify further.

\(0 < \left| {\left( {2x - 5} \right) - 3} \right| < \varepsilon \)

Therefore, it is proved that\(\mathop {\lim }\limits_{x \to 4} \left( {2x - 5} \right) = 3\).

Draw the graph of the function \(f\left( x \right) = 2x - 5\) locating the distance \(\delta \) and \(\varepsilon \) limits.