We have to compute the limit
[tex]\displaystyle \lim_{h\to 0} \frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}[/tex]
We want to use the cube difference formula:
[tex]a^3-b^3 = (a - b)(a^2 + ab + b^2)[/tex]
So, we can multiply both numerator and denominator as follows
[tex]\dfrac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h} = \dfrac{\left(\sqrt[3]{x+h}-\sqrt[3]{x}\right)\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)}{h\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)}[/tex]
The numerator is now the difference of cubes, so we have
[tex]\dfrac{x+h-x}{h\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)} = \dfrac{h}{h\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)} = \dfrac{1}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}[/tex]
As h approaches zero, we have
[tex]\displaystyle \lim_{h\to 0}\dfrac{1}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}} = \dfrac{1}{\sqrt[3]{(x+0)^2}+\sqrt[3]{x(x+0)}+\sqrt[3]{x^2}} = \dfrac{1}{3\sqrt{x^2}}[/tex]
