## Calculus: Early Transcendentals (2nd Edition)

The answer is $$\lim_{x\to2}\frac{\sqrt[3]{3x+2}-2}{x-2}=\frac{1}{4}.$$
To solve this limit follow the steps below. We will use the rule $a^3-b^3=(a-b)(a^2+ab+b^2)$ which gives $$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$ and thus $$\sqrt[3]{3x+2}-2=\frac{(\sqrt[3]{3x+2})^3-2^3}{(\sqrt[3]{3x+2})^2+2\sqrt[3]{3x+2}+2^2}=\frac{3x+2-8}{\sqrt[3]{(3x+2)^2}+2\sqrt[3]{3x+2}+4}=\frac{3x-6}{\sqrt[3]{(3x+2)^2}+2\sqrt[3]{3x+2}+4}=\frac{3(x-2)}{\sqrt[3]{(3x+2)^2}+2\sqrt[3]{3x+2}+4}.$$ Applying this to the given limit we get $$\lim_{x\to2}\frac{\sqrt[3]{3x+2}-2}{x-2}=\lim_{x\to2}\frac{\frac{3(x-2)}{\sqrt[3]{(3x+2)^2}+2\sqrt[3]{3x+2}+4}}{x-2}=\lim_{x\to2}\frac{3(x-2)}{(x-2)(\sqrt[3]{(3x+2)^2}+2\sqrt[3]{3x+2}+4)}=\lim_{x\to2}\frac{3}{\sqrt[3]{(3x+2)^2}+2\sqrt[3]{3x+2}+4}=\frac{3}{\sqrt[3]{(3\cdot2+2)^2}+2\sqrt[3]{3\cdot2+2}+4}=\frac{3}{\sqrt[3]{64}+2\sqrt[3]{8}+4}=\frac{1}{4}.$$