Prove that \(\lim\limits_{x \rightarrow 0} (x^3)= 8\)
Definitions and Theorems:
- Definition of functional limits
- [4.2.3] Sequential Criterion for Functional Limits
- [4.2.4] Algebraic Limit Theorem for Functional Limits
- [4.2.5] Divergence Criterion for Functional Limits
Plan
Let \(\epsilon > 0\), then we want to show that there exists a \(\delta > 0\) such that when
$$
\begin{align*}
|x| < \delta
\end{align*}
$$
then we have
$$
\begin{align*}
|x^3| < \epsilon
\end{align*}
$$
We want to choose \(\delta\) which controls how close \(x\) gets \(0\). We want to choose it such that it ensures that \(x^3\) is close to \(0\). So we can work backward and start from the conclusion to derive what \(\delta\) needs to be.
$$
\begin{align*}
|x^3| &< \epsilon \\
|x| &< \sqrt[3]{\epsilon}
\end{align*}
$$
so as long as we bound \(|x|\) by \(\delta = \sqrt[3]{\epsilon}\), then we can bound the functional limit.
Proof
Let \(\epsilon > 0\) be arbitrary. Let \(\delta = \sqrt[3]{\epsilon}\). Now, suppose that
$$
\begin{align*}
|x| < \delta = \sqrt[3]{\epsilon}
\end{align*}
$$
Observe now that
$$
\begin{align*}
|x^3| &= |x||x||x| < (\sqrt[3]{\epsilon})^3 = \epsilon
\end{align*}
$$
Thus, \(|x^3 - 0| < \epsilon\) as desired. \(\blacksquare\)
References
- Problem Statement Source: Aleph 0