Find limit (type 0/0) I'm struggling to find the limit 
$$I = \lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}$$
What I was trying:
$$ I = \lim_{x\to 0}\frac{1-x + 2\sqrt{1-x} + 1 - (1-x) - 1 - \sqrt[3]{8-x}}{x} $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x)- \sqrt[3]{8-x}}{x} \qquad \quad  $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x+ \sqrt[3]{8-x})}{x} \qquad \qquad  $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2}{x} - \lim_{x\to 0}\frac{(2-x+ \sqrt[3]{8-x})}{x}  \qquad  $$
Thank all of you for your answers.

Just use Cauchy's rule...
$$
\lim_{x \to 0} \dfrac{2\sqrt{1-x}-\sqrt[3]{8-x}}{x}=
\lim_{x\to 0} \left(-\frac{1}{\sqrt{1-x}}+\frac{1}{3(8-x)^{2/3}} \right)= -1 + \frac{1}{12}=-\frac{11}{12}
$$
