Epsilon Delta proof for $\lim_{x\to a} x^{1/3}$

Question

Is this proof for limit $\lim_{x\to a} x^{1/3}$ valid?

Lets first find Right Hand limit. $x-a > 0$. Let $x-a < \delta$ so that $x^{1/3}-a^{1/3} < \frac{\delta}{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}} < \frac{\delta}{a^{2/3}}$

So choosing $\delta = a^{2/3} \varepsilon$, we have $x^{1/3}-a^{1/3} \lt \varepsilon$ whenever $x-a \lt \delta$. Does this complete the proof?

Also I have read that we can choose $\delta$ bigger than what I chose. What does this statement mean? And why can we choose it?

Answer 1

If $a=0$, then for any $\epsilon>0$

$$|x^{1/3}-a^{1/3}|=|x^{1/3}|<\epsilon$$

whenever $|x|<\delta=\epsilon^3$.

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If $a> 0$, we first take $x$ such that $\frac{a}{2}<x<\frac{3a}{2}$.

Then, for any $\epsilon>0$

$$\begin{align} |x^{1/3}-a^{1/3}|&=\frac{|x-a|}{|x^{2/3}+a^{1/3}x^{1/3}+a^{2/3}|}\\\\ &<\frac{|x-a|}{a^{2/3}(1+2^{-1/3}+4^{-1/3})}\\\\ &<\epsilon \end{align}$$

whenever $|x-a|<\delta=\min\left(\frac a2,a^{2/3}(1+2^{-1/3}+4^{-1/3})\,\epsilon\right)$.

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The case $a<0$ is left as an exercise.