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 $\displaystyle\lim_{h→0} \dfrac {f(2+h)-f(2)}{h}$
 A: Have you tried multiplying top and bottom of $\frac{\sqrt[3]{2+h}-\sqrt[3]{2}}{h}$ by $(2+h)^{\frac{2}{3}}+\sqrt[3]{2}\cdot\sqrt[3]{2+h}+2^{\frac{2}{3}}$?
A: The limit $3\sqrt[3]4$  is given by the problem and it is easy to calculate. What the problem asks is to use the definition of limit to show that this limit is indeed  $3\sqrt[3]4$. This presents some difficulty.
Let $\epsilon\gt0$ be given. We need to find out $\delta \gt0$ such that $|x|\le \delta\Rightarrow |3\sqrt[3]4-f(x)|\lt\epsilon$.
We’ll prove that $\boxed{\delta =f(-\epsilon)-f(\epsilon)}$ fits.
Let us first notice that $f$ is strictly decreasing in the neighborhood of $0$ (derivative negative).
Proposition.- Let us $f(x)=\dfrac{\sqrt[3]{2+x}-\sqrt[3]2}{x}$ then for $x\gt0$ and near zero we have $0\lt f(-x)-f(x)\lt x$.
Proof.- (This is graphically verified with any good calculator, for example Desmos). We have
$$0\lt f(-x)-f(x)=\dfrac{\sqrt[3]{2-x}-\sqrt[3]2}{-x}-\dfrac{\sqrt[3]{2+x}-\sqrt[3]2}{x}=\frac{2\sqrt[3]2-\sqrt[3]{2-x}-\sqrt[3]{2+x}}{x} \\0\lt f(-x)-f(x)=\frac{\sqrt[3]2}{x}\left(2-\sqrt[3]{1-\frac x2}-\sqrt[3]{1+\frac x2}\right)$$
Using Taylor series $$\sqrt[3]{1-t}=1-\frac t3-\frac{t^2}{9}-\frac{5t^3}{81}-\frac{10t^4}{243}+O(t^5)\\\sqrt[3]{1+t}=1+\frac t3-\frac{t^2}{9}+\frac{5t^3}{81}-\frac{10t^4}{243}+O(t^5)$$ It follows
$$ f(-x)-f(x)=\frac{\sqrt[3]2}{x}\left(\frac29(\frac x2)^2+\frac{10}{243}(\frac x2)^4+O((\frac x2)^6)\right)=\frac{\sqrt[3]2}{x}\left(\frac{x^2}{18}+\frac{5x^4}{1944}+O\left(\frac{x^6}{64}\right)\right)$$ $$f(-x)-f(x)= x\left(\sqrt[3]2\left(\frac{1}{18}+\frac{5x^2}{1944}+\frac{1}{x^2}O\left(\frac{x^6}{64}\right)\right)\right)=x\left(\frac{1}{9\sqrt[3]4}+\frac{5x^2}{972\sqrt[3]4}+\frac{\sqrt[3]2}{x^2}O\left(\frac{x^6}{64}\right)\right)\lt x$$
because near zero the factor of $x$ is clearly less than $1$.
Now the rest of the proof is immediate. By construction, being $y_0=3\sqrt[3]4$, if $|x|\lt\delta=f(-\epsilon) -f(\epsilon)$ we have $$|y_0-f(x)|\lt f(-\epsilon) -f(\epsilon)\lt\epsilon$$ See the attached figure for clarity.

