Cube root of x by the Delta Method

$f\left(x\right)=\sqrt[3]{x}$

$\lim _{x\rightarrow 0}\ \frac{\sqrt[3]{x+\Delta x}-\sqrt[3]{x}}{\Delta x}$

Recall:

$a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$

Rearrange this formula:

$\frac{a^3−b^3}{(a^2+ab+b^2)}=a−b$

Set a as:

$a=\sqrt[3]{x+\Delta x}$

Set b as:

$b=\sqrt[3]{x}$

Substitute these new a and b values into the rearranged formula:

$\frac{\frac{x+\Delta x-x}{\left(\sqrt[3]{x+\Delta x}\right)^2+\left(\sqrt[3]{x+\Delta x}\cdot \sqrt[3]{x}\right)+\left(\sqrt[3]{x}\right)^2}}{\Delta x}$

Simplify:

$\frac{\Delta x}{\left[\left(\sqrt[3]{x+\Delta x}\right)^2+\left(\sqrt[3]{x+\Delta x}\cdot \sqrt[3]{x}\right)+\left(\sqrt[3]{x}\right)^2\right]\Delta x}$

Simplify more:

$\frac{1}{\left[\left(\sqrt[3]{x+\Delta x}\right)^2+\left(\sqrt[3]{x+\Delta x}\cdot \sqrt[3]{x}\right)+\left(\sqrt[3]{x}\right)^2\right]}$

Now take the bloody limit:

$=\frac{1}{\left(\sqrt[3]{x}\right)^2+\left(\sqrt[3]{x}\cdot \sqrt[3]{x}\right)+\left(\sqrt[3]{x}\right)^2}$

$=\frac{1}{3\left(\sqrt[3]{x}\right)^2}$

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Cube root of x by the Delta Method

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