How can i calculate this limite without using Hopital's rule?

Question

To Study the Differentiability of the this function at 3: $$g(x) = \frac{ x - 3 }{\,\rule{0pt}{4mm} \sqrt[ 3\, ]{\, x + 5\, }\, - 2 } ~~ for ~~ x ~~ \ne ~ 3 , ~~~~~~ g(3) =12.$$ i need to evaluat this limite "without using Hopital's rule" $$ \displaystyle\lim_{x \to 3}\ \left( \displaystyle \frac{\displaystyle\frac{x - 3} {\rule{0pt}{4mm}\,\sqrt[3\,]{\,x + 5\,}\, - 2} - 12}{x - 3}\right) $$ i've tried several way like : multiplying by the conjugate,.. doesn't work.

Answer 1

Let $y=\sqrt[3]{x+5}.$ Then $x\to 3$ gives $y\to 2$ and $x-3=y^3-8.$ Thus $${g(x)-12\over x-3}={{y^3-8\over y-2}-12\over y^3-8} ={1\over y-2}-{12\over (y-2)(y^2+2y+4)}\\ ={y^2+2y-8\over (y-2)(y^2+2y+4)}={y+4\over y^2+2y+4}\underset{y\to 2}{\longrightarrow}{1\over 2}$$