Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 26

Show that the statement $\displaystyle\lim\limits_{x \to 0} x^3 = 0$ is correct using the $\varepsilon$, $\delta$ definition of limit.

Based from the defintion,


$
\begin{equation}
\begin{aligned}

\phantom{x} \text{if } & 0 < |x - a| < \delta
\qquad \text{ then } \qquad
|f(x) - L| < \varepsilon\\

\phantom{x} \text{if } & 0 < |x-0| < \delta
\qquad \text{ then } \qquad
|x^3-0| < \varepsilon\\

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{That is,}\\
& \phantom{x} & \text{ if } 0 < |x| < \delta \qquad \text{ then } \qquad |x^3| < \varepsilon\\
\end{aligned}
\end{equation}
$

Or, taking the cube root of both sides of the inequality $|x^3| < \varepsilon$ , we get...
$\quad \text{if } 0 < |x| < \delta \quad \text{ then } \quad |x| < \sqrt[3]{\varepsilon}$

The statement suggests that we should choose $\displaystyle \delta = \sqrt[3]{\varepsilon}$

By proving that the assumed value of $\delta = \sqrt[3]{\varepsilon}$ will fit the definition...



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x| < \delta \text{ then, }\\
& \phantom{x} &
|x^3| < \delta^3 = (\sqrt[3]{\varepsilon})^3 = \varepsilon

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \text{Thus, }\\
& \phantom{x} \quad\text{if } 0 < |x| < \delta \qquad \text{ then } \qquad |x^3| < \varepsilon\\
& \text{Therefore, by the definition of a limit}\\
& \phantom{x} \qquad \lim\limits_{x \to 0}x^3 = 0

\end{aligned}
\end{equation}
$