Resolva a equação com radicais $\sqrt[3]{3-x}=0$

$\lim_{x\to-1}\left(\frac{12x^3+12x^2}{x^4-x^2}\right)$

$\lim_{x\to1}\left(\frac{x^3-1}{4x^3-x-3}\right)$

$\lim_{x\to8}\left(\frac{x-8}{x^2-64}\right)$

$\lim_{x\to3}\left(\frac{x^3-27}{x^2-9}\right)$

$\lim_{x\to4}\left(\frac{2-\sqrt{x}}{4-x}\right)$

$\lim_{x\to2}\left(\frac{x^4-16}{x-2}\right)$

$\lim_{x\to-4}\left(\frac{x+4}{x^2-16}\right)$