Proving continuity of inverse cube function

[Eclair_de_XII]

Homework Statement

Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ be defined by $f(x)=\sqrt[3]{x}$. Prove that for all $\epsilon>0$, there is a $\delta>0$ such that for $a\in \mathbb{R}$, if $|x-a|<\delta$ for $x\in \mathbb{R}$, then $|\sqrt[3]{x}-\sqrt[3]{a}|<\epsilon$.

Relevant Equations

Difference of cubes formula: $x^3-a^3=(x-a)(x^2+xa+a^2)$
No theorems allowed; must show by basic epsilon-delta proof.

The proof is given in two steps
1. Prove the lemma.
2. Use lemma to prove result.

%%1-Lemma%%
Assume $a\neq0$. Define $g:(-(|a|+1),|a|+1)\longrightarrow \mathbb{R}$ by $g(x)=\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2}$. Then $g$ is bounded from below by some positive number $m$.
---proof---
\begin{align}
g(x)&=&\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2}\\
&=&\sqrt[3]{x^2}+\sqrt[3]{xa}+\frac{1}{4}\sqrt[3]{a^2}+\frac{3}{4}\sqrt[3]{a^2}\\
&=&\left(\sqrt[3]{x}+\frac{1}{2}\sqrt[3]{a}\right)^2+\frac{3}{4}\sqrt[3]{a^2}\\
&\geq&\frac{3}{4}\sqrt[3]{a^2}\\
&=&m
\end{align}

%%2-Attempt at Proof%%
Let $\epsilon>0$ and set $\delta\leq\min\{1,\frac{\epsilon}{m}\}$. For $a\neq0$, if $|x-a|=|\sqrt[3]x-\sqrt[3]a|\left|\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2}\right|<\delta$, then:

\begin{align}
1&>&|x-a|\\
&\geq&||x|-|a||\\
&\geq&|x|-|a|
\end{align}

This implies that $|x|<1+|a|$, which gives the necessary condition for the lemma.

\begin{align}
\left|\sqrt[3]x-\sqrt[3]a\right|&=&\frac{|x-a|}{\left|\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2}\right|}\\
&=&\frac{|x-a|}{\left|g(x)\right|}\\
&\leq&\frac{|x-a|}{m}\\
&<&\delta\cdot m\\
&\leq&\frac{\epsilon}{m}\cdot m\\
&<&\epsilon
\end{align}

If $a=0$, then set $\delta=\epsilon^3$ so that if $|x|<\delta$, then $\left|\sqrt[3]x\right|<\delta^{\frac{1}{3}}=(\epsilon^3)^\frac{1}{3}=\epsilon$.

 

[Hall]

Couldn't we simply take $|x-a| \lt 1$ and then get a bound for $|x^2 + xa + a^2|$?

 

[Mark44]

Couldn't we simply take $|x-a| \lt 1$ and then get a bound for $|x^2 + xa + a^2|$?

This is the usual approach for this and similar proofs.