<s>[INST] limit $\lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2}$ Can you help me with 
$$\lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2}$$
I've already tried to multiply at conjugate expression but I failed.
 [/INST] $$  \lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2}  \\$$
Let $ a= a=\sqrt[3]{x-6} $ and $ b=  {2} $ . Then $ a^3+b^3 = (a+b)(a^2-ab+b^2)$ . So $ a^3+b^3 = x+2 $  and $ a+b = \sqrt[3]{x - 6}+2 $  . So $ \frac{\sqrt[3]{x - 6}+2}{x + 2} $ = $  \frac{a+b}{a^3+b^3}$ = $\frac{1}{ a^2 - ab +b^2} $ . Hope so this helps a lot . 
Here at $ x =-2 $ , the value of a is -2 . So at $  x= -2 $ the value of $ a^2-ab+b^2 = 4-(-2)2+4 = 12 $ . 
So the ultimate value of  $$  \lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2}  $$ is $ \frac{1}{12} $
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