Find limit of a fraction without derivatives: $\lim\limits_{h\rightarrow0} \frac{\sqrt[3]{x+h} - \sqrt[3]{x}}h$ How to find limit of this?
\begin{equation}
\lim_{h \rightarrow 0} \frac {\sqrt[3]{x+h} - \sqrt[3]{x}} {h}.
\end{equation}
I see here the derivative definition, so I understand in "derivative mind" that answer has to be \begin{equation} \frac{1}{3} x^{-\frac{2}{3}}, \end{equation}but in the book this expression goes before derivatives, so it must be another approach to find this limit.
 A: Hint: $$(a-b)(a^2+ab+b^2)=a^3-b^3.$$
A: Rationalise the numerator by the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$
$\lim _{h\rightarrow 0}{\frac {\sqrt [3]{x+h}-\sqrt [3]{x}}{h}}$
$=\lim _{h\rightarrow 0}{\frac {\sqrt [3]{x+h}-\sqrt [3]{x}}{h}}{\frac { \left( x+h \right) ^{2/3}+\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3
}}{ \left( x+h \right) ^{2/3}+\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3}}}
$
$=\lim _{h\rightarrow 0}{\frac {(x+h)-x}{h \left(  \left( x+h \right) ^{2/3}+
\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3} \right) }}$
$=\lim _{h\rightarrow 0}{\frac {1}{   \left( x+h \right) ^{2/3}+
\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3}  }}$
$={\frac {1}{   \left( x \right) ^{2/3}+
\sqrt [3]{x}\sqrt [3]{x}+{x}^{2/3}  }}$ by plugging in $h=0$
$={\frac{1}{3x^{2/3}}}$
=$\frac{1}{3}x^{-2/3}$
A: Hint
Another one (assuming $x\neq 0$) :$$\sqrt[3]{x+h}=x^{\frac{1}{3}}\sqrt[3]{1+\frac{h}{x}}$$ Now, if you remember that, when $y$ is small, $(1+y)^a \simeq 1+ ay$, then now, replace $y$ by $\frac{h}{x}$ and so $$ \sqrt[3]{x+h}=x^{\frac{1}{3}}\sqrt[3]{1+\frac{h}{x}}\simeq x^{\frac{1}{3}}(1+\frac{h}{3x})=x^{\frac{1}{3}}+\frac{h}{3x^{\frac{2}{3}}}$$
I am sure that you can take from here.
A: HINT:
Set $\displaystyle\sqrt[3]{x+h}=a\implies x+h=a^3,\sqrt[3]x=b\implies x=b^3; a^3-b^3=h$
As $\displaystyle h\to0, a\to  b\implies a\ne b$
Finally, the limit should be $\displaystyle3/b^2=3/(x^{\dfrac23})$
