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I am trying to prove that a function is continuous at a point a using the $\epsilon$-$\delta$ theorem. I managed to find a $\delta$ in this case $|2x^2+1 - (2a^2+1)| < \epsilon$. But I have a hard time when the function under consideration is $f(x) = \sqrt[3]{x}$. That is, I want to have if $|x-a|<\delta$, then $|\sqrt[3]{x}-\sqrt[3]{a}| < \epsilon$. Any suggestions?

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2 Answers 2

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$$|\sqrt[3]{x}-\sqrt[3]{a}| = |\sqrt[3]{x}-\sqrt[3]{a}| \times \frac {| x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} = \frac {|x - a|}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} \le \frac {| x - a |}{| {ax} |^{\frac 1 3}}$$

The final inequality is due to the fact that $x^{\frac 2 3} + a^{\frac 2 3} \ge 0$.

Let us assume $a \neq 0$. Then We can bound $|x|$ as follows.

Say $|x - a| \lt |a| $ then $ |x| \lt 2|a| \implies |ax| \lt 2|a|^2 \implies \frac {1}{2|a|^2} \lt \frac {1} {|ax|}$.

Therefore $|\sqrt[3]{x}-\sqrt[3]{a}| \lt \frac {| x - a |}{| {a} |^{\frac 2 3}}$ as long as $|x - a| \lt |a|$ and $a \neq 0$. So if we pick $\delta = \text {Min} \{ |a|, \epsilon|a|^{\frac 2 3} \}$ then $|x - a| \lt \delta \implies |\sqrt[3]{x}-\sqrt[3]{a}| \lt \epsilon$. This proves the function is continuous everywhere except at $0$. To get rid of the case when $a = 0$ just pick $\delta = \epsilon^{3} $

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    $\begingroup$ Or if you want to get fancy, you can show $$|\sqrt[k]x-\sqrt[k]a|\le\sqrt[k]{|x-a|}$$ for any $x,a\ge0$. See (math.stackexchange.com/questions/678226/#678332) $\endgroup$ Feb 23, 2014 at 19:24
  • $\begingroup$ Thanks Ishfaaq, very good answer. $\endgroup$
    – user29163
    Feb 23, 2014 at 19:53
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    $\begingroup$ user29163: Anytime. Although I advise we should both look into @LutzL's suggestion above.. $\endgroup$
    – Ishfaaq
    Feb 23, 2014 at 19:56
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    $\begingroup$ @Ishfaaq Just found your nice effort here as I am working on showing $f(x)=x^{1/3}$ is continuous at any real number $a$. However, I think there might be an error here. I agree with your effort to get $$\frac{1}{2|a|^2}<\frac{1}{|ax|}$$, which means that $$\frac{1}{2^{1/3}|a|^{2/3}}<\frac{1}{|ax|^{1/3}},$$ but note that $$\frac{|x-a|}{|ax|^{1/3}}=|x-a|\cdot\frac{1}{|ax|^{1/3}}>|x-a|\cdot\frac{1}{2^{1/3}|a|^{2/3}}.$$ $\endgroup$
    – David
    Feb 17, 2016 at 18:12
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Let $\epsilon>0$ be given.

Case 1: Let $a>0$. Then we have \begin{align*} |f(x)-f(a)|=\left|x^{1/3}-a^{1/3}\right|\frac{\left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right|}{\left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right|}=\frac{|x-a|}{\left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right|} \end{align*} Now consider $x\in \mathbb R$ such that $x>\frac{a}{2}$. Then $x^{1/3}>0$ and $a^{1/3}>0$. Clearly $x^{2/3}, a^{2/3}>0$. $$ \left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right| = x^{2/3}+a^{2/3}+x^{1/3}a^{1/3} >a^{2/3} $$

Case 2: Let $a<0$. Now consider $x\in \mathbb R$ such that $x<\frac{a}{2}$. Then $x^{1/3}<0$ and $a^{1/3}<0$. Clearly $x^{2/3}, a^{2/3}>0$. $$ \left|x^{2/3}+a^{2/3}+x^{1/3}a^{1/3}\right| = x^{2/3}+a^{2/3}+x^{1/3}a^{1/3} > a^{2/3} $$

Then we have $$ |f(x)-f(a)|< \frac{|x-a|}{a^{2/3}} $$ Let us choose $\delta=\min \left\{a^{2/3}\epsilon, \dfrac{|a|}{2}\right\}$ and consider $x\in \mathbb R$ such that $|x-a|<\delta$. Then we have $$ |f(x)-f(a)|<\epsilon $$

Case 3: Suppose $a=0$, then choose $\delta=\epsilon^3$.
$$ |f(x)-f(0)|=|x^{1/3}|<\delta^{1/3}=\epsilon $$ Hence $f$ is continuous on $\mathbb R$.

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