Question

Given that \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{2}}} \left( {{\bf{5}}x - {\bf{7}}} \right) = {\bf{3}}\),illustrate Definition 2 by finding values of \(\delta \) that corresponds to \(\varepsilon = {\bf{0}}.{\bf{1}}\), \(\varepsilon = {\bf{0}}.{\bf{05}}\), and \(\varepsilon = {\bf{0}}.{\bf{01}}\).

Short answer

For \(\varepsilon = 0.1\), \(\delta = 0.02\).

For \(\varepsilon = 0.05\), \(\delta = 0.01\).

For \(\varepsilon = 0.01\), \(\delta = 0.002\)

Step-by-step solution

Step 1:Solve the expression \(\left| {\left( {{\bf{5}}x - {\bf{7}}} \right) - {\bf{3}}} \right|\)

Solve the expression\(\left| {\left( {5x - 7} \right) - 3} \right|\) as shown below:

\(\begin{aligned}\left| {\left( {5x - 7} \right) - 3} \right| = \left| {5x - 10} \right|\\ = 5\left| {x - 2} \right|\end{aligned}\)

Use definition 2 for the value of \(\delta \)

Bydefinition 2:

\(\begin{aligned}\left| {f\left( x \right) - L} \right| < \varepsilon \\5\left| {x - 2} \right| < \varepsilon \\\left| {x - 2} \right| < \frac{\varepsilon }{5}\end{aligned}\)

Find the value of \(\delta \)for the given values of \(\varepsilon \)

For \(\varepsilon = 0.1\), the number is shown below:

\(\begin{aligned}\left| {x - 2} \right| < \frac{{0.1}}{5}\\ < 0.02\end{aligned}\)

For \(\varepsilon = 0.05\), the number is shown below:

\(\begin{aligned}\left| {x - 2} \right| < \frac{{0.05}}{5}\\ < 0.01\end{aligned}\)

For \(\varepsilon = 0.01\), the number is shown below:

\(\begin{aligned}\left| {x - 2} \right| < \frac{{0.01}}{5}\\ < 0.002\end{aligned}\)

So, the values of \(\delta \) are: for \(\varepsilon = 0.1\), \(\delta = 0.02\), for \(\varepsilon = 0.05\), \(\delta = 0.01\), and for \(\varepsilon = 0.01\), \(\delta = 0.002\).