<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] $f(x)=2(1-x)^{1/2}$;
$g(x)=\sqrt[3]{8-x}$.
$F(x)=\dfrac{f(x)-f(0)}{x}=$
$\dfrac{2\sqrt{1-x}-2}{x}$
$G(x)=\dfrac{g(x)-g(0)}{x}=\dfrac{\sqrt[3]{8-x}-2}{x}$
$F'(0)= [-(1-x)^{-1/2}]_{x=0}=-1$.
$G'(0)=[-(1/3)(8-x)^{-2/3}]_{x=0}=-(1/3)(1/4)=-1/(12).$
Hence : $F'(0)-G'(0)= -11/(12)$.
Recall:
$\lim_{ x \rightarrow 0}\dfrac{F(x)-F(0)}{x}=F'(0)$, 
similarly for $G$.
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