Question

(a)For the limit \(\mathop {\lim }\limits_{x \to 1} \left( {{x^3} + x + 1} \right) = 3\), use a graph to find a value of \(\delta \) that corresponds to \(\varepsilon = 0.4\).

(b)By using the computer algebra system to solve the cubic equation \({x^3} + x + 1 = 3 + \varepsilon \), find the largest possible value of \(\delta \) that works for any given \(\varepsilon > 0\).

(c)Put \(\varepsilon = 0.4\) in your answer to part (b) and compare with your answer to part (a).

Short answer

(a)The value of \(\delta \) corresponds to \(\varepsilon > 0.4\) is \(0.093\).

(b)The largest possible value of \(\delta \) that works for any given \(\varepsilon > 0\) is \(\delta = x\left( \varepsilon \right) - 1\).

(c)Answer which obtained in part (c) is same as the answer or part (a).

Step-by-step solution

(a)Step 1: finding a value of \(\delta \) that corresponds to \(\varepsilon  = 0.4\)

Here given that,

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

So, by definition of limit,

\(\left| {x - 1} \right| < \delta \Rightarrow \left| {{x^3} + x + 1} \right| < 0.4\)

Now, graph of the function is given by,

Also, we have given as\(\mathop {\lim }\limits_{x \to 1} \left( {{x^3} + x + 1} \right) = 3\)

So, we consider region around point\(\left( {1,\,3} \right)\)

Now, rewrite the inequality as,

\(\begin{array}{c} - 0.4 < {x^3} + x + 1 - 3 < 0.4\\ - 0.4 + 3 < {x^3} + x + 1 < 0.4 + 3\\2.6 < {x^3} + x + 1 < 3.4\end{array}\)

Here, we have to determine the value of \(x\) for which the graph \(y = {x^3} + x + 1\) lies between the lines \(y = 2.6\) and \(y = 3.4\). Near \(\left( {1,\,3} \right)\)

From above graph we can see that at\(y = 2.6\)we get\(x = 0.89\)and for\(y = 3.4\)we get\(x = 1.09\).

So, by definition,

\(\begin{array}{c}\left| {x - 1} \right| < \delta \\\left| {0.89 - 1} \right| < \delta \\\left| {0.109} \right| < \delta \end{array}\)

Or,

\(\begin{array}{c}\left| {x - 1} \right| < \delta \\\left| {1.093 - 1} \right| < \delta \\\left| {0.093} \right| < \delta \end{array}\)

By definition of limit:

\(\begin{array}{c}\delta = \min \left\{ {0.109,\,0.093} \right\}\\ = 0.093\end{array}\)

Hence, the value of \(\delta \) corresponds to \(\varepsilon > 0.4\) is \(0.093\).

(b)Step 2:  find the largest possible value of \(\delta \) that works for any given \(\varepsilon  > 0\) by using the computer algebra system to solve the cubic equation \({x^3} + x + 1 = 3 + \varepsilon \)

Here we have,

\(\begin{aligned}{x^3} + x + 1 = 3 + \varepsilon \\\varepsilon = {x^3} + x - 2\end{aligned}\)

Now, by using computer algebra system to solve above cubic equation, we get the solution as:

\(x\left( \varepsilon \right)=\frac{\sqrt[3]{\sqrt{3}\sqrt{27{{\varepsilon }^{2}}+108\varepsilon +112}+9\varepsilon +18}}{\sqrt[3]{2}{{3}^{{2}/{3}\;}}}-\frac{\sqrt[3]{{}^{2}/{}_{3}}}{\sqrt[3]{\sqrt{3}\sqrt{27{{\varepsilon }^{2}}+108\varepsilon +112}+9\varepsilon +18}}\)

Now, we have to take \(\delta \) as \(\left| {x - 1} \right| < \delta \)

So, \(\delta = x\left( \varepsilon \right) - 1\).

Hence, the largest possible value of

\(\delta \) that works for any given \(\varepsilon > 0\) is \(\delta = x\left( \varepsilon \right) - 1\).

(c)Step 3: Put \(\varepsilon  = 0.4\) in answer to part (b) and comparing with your answer to part (a)

In step 2, we obtained that

\(\delta = x\left( \varepsilon \right) - 1\) where,

\(x\left( \varepsilon \right)=\frac{\sqrt[3]{\sqrt{3}\sqrt{27{{\varepsilon }^{2}}+108\varepsilon +112}+9\varepsilon +18}}{\sqrt[3]{2}{{3}^{{2}/{3}\;}}}-\frac{\sqrt[3]{{}^{2}/{}_{3}}}{\sqrt[3]{\sqrt{3}\sqrt{27{{\varepsilon }^{2}}+108\varepsilon +112}+9\varepsilon +18}}\)

Thus, we get value of \(\delta \) as:

\(\delta =\frac{\sqrt[3]{\sqrt{3}\sqrt{27{{\varepsilon }^{2}}+108\varepsilon +112}+9\varepsilon +18}}{\sqrt[3]{2}{{3}^{{2}/{3}\;}}}-\frac{\sqrt[3]{{}^{2}/{}_{3}}}{\sqrt[3]{\sqrt{3}\sqrt{27{{\varepsilon }^{2}}+108\varepsilon +112}+9\varepsilon +18}}-1\)

By substituting \(\varepsilon = 0.4\) in above equation,

\(\begin{align} & \delta =\frac{\sqrt[3]{\sqrt{3}\sqrt{27{{\left( 0.4 \right)}^{2}}+108\left( 0.4 \right)+112}+9\left( 0.4 \right)+18}}{\sqrt[3]{2}{{3}^{{2}/{3}\;}}}-\frac{\sqrt[3]{{}^{2}/{}_{3}}}{\sqrt[3]{\sqrt{3}\sqrt{27{{\left( 0.4 \right)}^{2}}+108\left( 0.4 \right)+112}+9\left( 0.4 \right)+18}}-1 \\ & =1.093-1 \\ & =0.093 \end{align}\)

In part (a), we obtained that \(\delta = 0.093\) by using the graph.

Here we obtained \(\delta = 0.093\) analytically.

Hence, the answer which obtained in part (c) is same as the answer or part (a).\(\)