Use of definition of limit to prove $\displaystyle\lim_{x→2} \sqrt[3]{x}= \sqrt[3]{2}$

Question

I know that by definition I have to prove that $$\displaystyle\lim_{x→2} \sqrt[3]{x}= \sqrt[3]{2}⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹|\sqrt[3]{x}-2|<ϵ.$$

I have this:

$ |\sqrt[3]{x}-2|= |\dfrac{(\sqrt[3]{x}-2)((\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)}{(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2}|= \dfrac{|x-2|}{|(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)|}$.

I know that |x-2|<$\delta$. But I don't know what $\delta$ choose and how narrow down the denominator.

Answer 1

If $x >0$ we have $ \dfrac{|x-2|}{|(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)|} \leq \frac {|x-2|} {2^{2/3}}$ so you can take $\delta=\min \{1,\epsilon {2^{2/3}}\}$. [Note that $x>0$ for $|x-2| <\delta$].

Answer 2

(Note that $a^2+ab+b^2=(a+\frac b2)^2+\frac 34 b^2>\frac 34 b^2$.)

$$\dfrac{|x-2|}{|(\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{2}+(\sqrt[3]{2})^2)|} = \frac{|x-2|}{\left(\sqrt[3]{x}+\frac{\sqrt[3]{2}}{2}\right)^2 + \frac 34 (\sqrt[3]{2})^2 } < \frac{|x-2|}{0+\frac 12 \cdot 1}= 2|x-2|$$

So $|x-2| < \delta: = \frac{\varepsilon}{2} \implies |\sqrt[3]{x}-\sqrt[3]{2}| < \varepsilon$.