Using factorization backwards

A common situation occurring in several limit calculations uses the factorization of differences of powers to rationalize numerators in fractions involving roots:

From $a^2-b^2=(a-b)(a+b)$ we get

$$\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}\f... ...\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\frac{h}{h(\sqrt{x+h}+\sqrt{x}}.$$

From $a^3-b^3=(a-b)(a^2+ab+b^2)$ we get

$$\begin{array}{ccl}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h} &=&\fra... ...h(\sqrt[3]{(x+h)^2}+\sqrt[3]{(x+h)x}+\sqrt[3]{x^2})}\end{array}$$