- #1
Eclair_de_XII
- 1,083
- 91
- 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##.
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##.
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