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.

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 $h=0$
$={\frac{1}{3x^{2/3}}}$
=$\frac{1}{3}x^{-2/3}$
