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.

If you already know the concept of the first derivative you can do the following:
$$\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}= \frac{2\sqrt{1-x} - 2}{x} - \frac{\sqrt[3]{8-x}-2}{x} \stackrel{x\to 0}{\longrightarrow} \left.\frac{d}{dx}\left(2\sqrt{1-x}- \sqrt[3]{8-x}\right) \right|_{x=0}$$
Differentiating and setting $x= 0$ gives $-\frac{11}{12}$.
