Question

For the limit

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{x} = 2\)

illustrate Definition 2 by finding values of \(\delta \) that correspond

to \(\varepsilon = 0.5\) and \(\varepsilon = 0.1\)

Short answer

• For \(\varepsilon = 0.5\), the value of \(\delta \) is \(0.215\).
• For \(\varepsilon = 0.1\), the value of \(\delta \) is \(0.048\).

Step-by-step solution

Plot the graph of the function.

Draw the graph of the function\(f\left( x \right) = \frac{{{e^{2x}} - 1}}{x}\)by using the graphing calculator as:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\frac{{{e^{2X}} - 1}}{X}\)in the\({Y_1}\)tab.

2. Enter the “GRAPH” button in the graphing calculator.

The graph of the function \(f\left( x \right) = \frac{{{e^{2x}} - 1}}{x}\) is shown below:

Solve the condition

From the given limit and consider \( \in = 0.5\), we have \(\left| {\frac{{{e^{2x}} - 1}}{x} - 2} \right| < 0.5\).

Apply the absolute property, that is, if \(\left| x \right| < a\) then \( - a < x < a\).

Rewrite the condition \(\left| {\frac{{{e^{2x}} - 1}}{x} - 2} \right| < 0.5\) as:

\(\begin{aligned} - 0.5 < \frac{{{e^{2x}} - 1}}{x} - 2 < 0.5\\1.5 < \frac{{{e^{2x}} - 1}}{x} < 2.5\\ - 0.303 < x < 0.215\end{aligned}\)

Observe the graph and obtain the value of \({\delta _1}\) and \({\delta _2}\)

From the graph, it is observed that \( - {\delta _1} \approx - 0.303\) and \({\delta _2} \approx 0.215\).

Solve the obtained condition.

\(\begin{aligned} - {\delta _1} \approx - 0.303\\{\delta _1} \approx 0.303\end{aligned}\)

Obtain the value of \(\delta \)

The value of \(\delta \) is the minimum value of \({\delta _1}\), and \({\delta _2}\).

\(\begin{aligned}\delta & = {\rm{min}}\left\{ {0.303,0.215} \right\}\\ & = 0.215\end{aligned}\)

Therefore, the value of \(\delta \) is \(0.215\) for \(\varepsilon = 0.5\).

Solve the condition

From the given limit and consider \(\varepsilon = 0.1\), we have \(\left| {\frac{{{e^{2x}} - 1}}{x} - 2} \right| < 0.1\).

Apply the absolute property, that is, if \(\left| x \right| < a\) then \( - a < x < a\).

Rewrite the condition \(\left| {\left( {{x^3} - 3x + 4} \right) - 6} \right| < 0.1\) as:

\(\begin{aligned} - 0.1 < \frac{{{e^{2x}} - 1}}{x} - 2 < 0.1\\1.9 < \frac{{{e^{2x}} - 1}}{x} < 2.1\\ - 0.052 < x < 0.048\end{aligned}\)

Observe the graph and obtain the value of \({\delta _1}\) and \({\delta _2}\)

From the graph, it is observed that \( - {\delta _1} \approx - 0.052\) and \({\delta _2} \approx 0.048\).

Solve the obtained condition.

\(\begin{aligned} - {\delta _1} \approx - 0.052\\{\delta _1} \approx 0.052\end{aligned}\)

Obtain the value of  \(\delta \)

The value of \(\delta \) is the minimum value of \({\delta _1}\), and \({\delta _2}\).

\(\begin{aligned}\delta & = {\rm{min}}\left\{ {0.052,0.048} \right\}\\ & = 0.048\end{aligned}\)

Therefore, the value of \(\delta \) is \(0.048\) for \(\varepsilon = 0.1\).