Question

Prove the statement using the\(\varepsilon \), \(\delta \)definition of a limit.

\(\mathop {\lim }\limits_{x \to 1} \frac{{2 + 4x}}{3} = 2\)

Short answer

It is proved that\(\mathop {\lim }\limits_{x \to 1} \frac{{2 + 4x}}{3} = 2\).

Step-by-step solution

Describe the given information

It is required to prove\(\mathop {\lim }\limits_{x \to 1} \frac{{2 + 4x}}{3} = 2\)by using\(\varepsilon \),\(\delta \)definition.

Prove that \(\mathop {\lim }\limits_{x \to 1} \frac{{2 + 4x}}{3} = 2\)

Consider the limit\(\mathop {\lim }\limits_{x \to 1} \frac{{2 + 4x}}{3} = 2\).

Use \(\varepsilon - \delta \) definition to prove the statement\(\mathop {\lim }\limits_{x \to 1} \frac{{2 + 4x}}{3} = 2\).

If \(0 < \left| {x - 1} \right| < \delta \)then\(0 < \left| {\left( {\frac{{2 + 4x}}{3}} \right) - 2} \right| < \varepsilon \).

Simplify in the absolute value inequality as follows.

If \(0 < \left| {x - 1} \right| < \delta \)then\(0 < \left| {2 + 4x - 6} \right| < \varepsilon \).

If \(0 < \left| {x - 1} \right| < \delta \)then\(0 < \left| {4x - 4} \right| < \varepsilon \).

If \(0 < \left| {x - 1} \right| < \delta \)then\(0 < \left| {4\left( {x - 1} \right)} \right| < \varepsilon \).

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

If \(0 < \left| {x - 1} \right| < \delta \)then\(0 < 4\left| {\left( {x - 1} \right)} \right| < \varepsilon \).

Divide the inequality by \(4\)

If \(0 < \left| {x - 1} \right| < \delta \)then\(0 < \left| {x - 1} \right| < \frac{\varepsilon }{4}\).

Choose\(\delta = \frac{\varepsilon }{4}\), and then multiply each side of the inequality by\(3\).

\(\begin{array}{c}0 < 4\left| {x - 1} \right| < \varepsilon \\0 < \left| {4\left( {x - 1} \right)} \right| < \varepsilon \\0 < \left| {4x - 4} \right| < \varepsilon \\0 < \left| {4x + 2 - 6} \right| < \varepsilon \end{array}\)

Simplify further.

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

Therefore, it is proved that\(\mathop {\lim }\limits_{x \to 1} \frac{{2 + 4x}}{3} = 2\).