I can't think of any change of variable to solve this limit, I can't use derivatives or series. Can you please guide me on how to do it? Thank you
$$\underset {x\to 6} {\text {lim}}\frac {x -\sqrt[3] {x + 2} - \sqrt {3 x - 2}} {x -\sqrt[3] {4 x + 3} - \sqrt {x + 3}}$$
Edit taking the suggestion of @Paramanand Singh making x=6+h , x->6 ,h->0 I arrive at the following ??
$$\underset {h\to 0} {\text {lim}}\frac {(h + 6) - \sqrt[3] {h + 6 + 2} - \sqrt {3 (h + 6) - 2}} {(h + 6) - \sqrt[3] {4 (h + 6) + 3} - \sqrt {h + 6 + 3}}$$
$$\underset {h\to 0 } {\text {lim}}\frac {(h + 6) - \sqrt[3] {h + 8} - \sqrt {3 h + 16}} {(h + 6) - \sqrt[3] {4 h + 27} - \sqrt {h + 9}}$$
$$\displaystyle \huge \lim_{h->0}\frac{\frac{(h+6)-\sqrt[3]{8+h}-\sqrt{3h+16}}{h}}{\frac{(h+6)-\sqrt[3]{4h+27}-\sqrt{h+9}}{h}}$$
Edit 2 : by manipulating the numerator and denominator, I get 4 limits and when I develop them individually I reach the desired result. Thanks to all
Well one way(which may not be satisfactory to you) is to expand the numerator and denominator about $x=6$. We get the following result for the numerator $$x-\sqrt{3x-2}-\sqrt[3]{x+2} = \frac{13}{24}(x-6)+O((x-6)^2)$$ and for the denominator $$x - \sqrt{x+3} - \sqrt[3]{4x+3} = \frac{37}{54}(x-6)+O((x-6)^2)$$ And therefore the result is after dividing and neglecting the higher order terms $\frac{117}{148}$.
For what it's worth, Wolframalpha gives the same result.