Question

19-32: Prove the statement using the \(\varepsilon ,\delta \) definition of a limit.

32. \(\mathop {\lim }\limits_{x \to 2} {x^3} = 8\)

Short answer

It is proved that \(\mathop {\lim }\limits_{x \to 2} {x^3} = 8\).

Step-by-step solution

Precise Definition of a limit

Consider \(f\) as afunctiondefined on someopen interval that contains the number \(a\), except possibly at \(a\) itself.

Then, we say that thelimit of \(f\left( x \right)\) as \(x\)approaches \(a\) is \(L\), and it can be written as\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = L\).

If for all numbers \(\varepsilon > 0\) there exists a number \(\delta > 0\) such that if \(0 < \left| {x - a} \right| < \delta \) then \(\left| {f\left( x \right) - L} \right| < \varepsilon \).

Prove the statement using the definition of a limit

Let \(\varepsilon > 0\)be a given positive number. We want \(\delta > 0\) such that when\(0 < \left| {x - 2} \right| < 1\) then \(\left| {{x^3} - 8} \right| < \varepsilon \).

In this case, \(\left| {{x^3} - 8} \right| = \left| {\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)} \right|\). When \(\left| {x - 2} \right| < 1\) that is \(1 < x < 3\) then, it can be represented as:

\(\begin{array}{c}{x^2} + 2x + 4 < {3^2} + 2\left( 3 \right) + 4\\ = 19\end{array}\)

Therefore, \(\left| {{x^3} - 8} \right| = \left| {x - 2} \right|\left( {{x^2} + 2x + 4} \right) < 19\left| {x - 2} \right|\).

Let us choose \(\delta = \min \left\{ {1,\frac{\varepsilon }{{19}}} \right\}\). If \(0 < \left| {x - 2} \right| < \delta \) then \(\left| {x - 2} \right| < \frac{\varepsilon }{{19}}\) and \(\left| {{x^2} + 2x + 4} \right| < 19\). Therefore,

\(\begin{aligned}\left| {{x^3} - 8} \right| &= \left| {x - 2} \right|\left( {{x^2} + 2x + 4} \right)\\ < \frac{\varepsilon }{{19}} \cdot 19\\ &= \varepsilon \end{aligned}\)

Hence, \(\mathop {\lim }\limits_{x \to 2} {x^3} = 8\) according to the definition of a limit.

Thus, it is proved that \(\mathop {\lim }\limits_{x \to 2} {x^3} = 8\).