<s>[INST] 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.
 [/INST] Until the last step your attempt is correct. However, the last step (splitting to two limits) is not allowed, because it becomes the indetermined form $\infty-\infty$. Here is an alternative way with splitting:
$$\lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}=\\\lim_{x\to 0}\frac{(2\sqrt{1-x}-2)+(2-\sqrt[3]{8-x})}{x}=\\
\lim_{x\to 0}\frac{2(\sqrt{1-x}-1)}{x}+\lim_{x\to 0}\frac{2-\sqrt[3]{8-x}}{x}=\\
\lim_{x\to 0}\frac{2(1-x-1)}{x(\sqrt{1-x}+1)}+\lim_{x\to 0}\frac{8-(8-x)}{x(4+2\sqrt[3]{8-x}+\sqrt[3]{(8-x)^2})}=\\
-1+\frac1{12}=-\frac{11}{12}.$$
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