Question

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

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

Short answer

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

Step-by-step solution

Describe the given information

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

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

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

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

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

Add the fraction in the absolute value on the right as follows.

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

This statement can also be written as,

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

Multiply throughout the later inequality by\(3\), so the above statement becomes,

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

Choose\(\delta = 3\varepsilon \), then

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

Simplify further.

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

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

Draw the graph of the function \(f\left( x \right) = \frac{1}{3}x + 1\) locating the distance \(\delta \) and \(\varepsilon \) limits.