<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] The key is to use the slightly less popular limit formula $$\lim_{t\to a} \frac{t^n-a^n} {t-a} =na^{n-1}\tag{1}$$ The expression under limit in question can be written as $$2\cdot \frac{\sqrt{1-x}-1}{x}-\frac{\sqrt[3]{8-x}-2}{x}$$ and using $u=1-x,v=8-x$ we can see that $u\to 1,v\to 8$ and the above expression can be written as $$\frac{v^{1/3}-8^{1/3}}{v-8}-2\cdot\frac{u^{1/2}-1^{1/2}}{u-1}$$ By formula $(1)$ the above tends to $$\frac{1}{3}\cdot 8^{-2/3}-2\cdot\frac{1}{2}\cdot 1^{-1/2}=-\frac{11}{12}$$ While doing any algebraic manipulation on the expression (whose limit is to be evaluated) the goal should not only be a simplification but also transforming it into a form amenable to the use of standard / well-known limits. 
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