<s>[INST] Use of definition of limit to prove $\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3\cdot \sqrt[3]{4} }$ I know that by definition I have to prove that $$\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3 \cdot \sqrt[3]{4}  }⟺∀ϵ>0,∃δ>0,\, 0<|h−0|<δ⟹| \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot \sqrt[3]{4}  } |<ϵ.$$
I have this:
$\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot  \sqrt[3]{4}} \cdot\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}} $
= $\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac{\sqrt[3]{2}}{6} $
=$ \dfrac{6\cdot \sqrt[3]{2+h}-6\cdot \sqrt [3]{2}-\sqrt[3]{2}\cdot h}{6h} $
=$ \left |\dfrac{1}{(\sqrt[3]{2+h})^2 + \sqrt[3]{2+h}\sqrt[3]{2}+(\sqrt[3]{2})^2 } -  \dfrac {\sqrt[3]{2}} {6}\right | $
=$   $
I don't know how continue for use the fact $|h|<\delta$
The original exercise says: $f(x)=\sqrt[3]{x}$. Found $\displaystyle\lim_{h→0} \dfrac {f(2+h)-f(2)}{h}$
 [/INST] Have you tried multiplying top and bottom of $\frac{\sqrt[3]{2+h}-\sqrt[3]{2}}{h}$ by $(2+h)^{\frac{2}{3}}+\sqrt[3]{2}\cdot\sqrt[3]{2+h}+2^{\frac{2}{3}}$?
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