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.
 A: I would substitute $$a=\sqrt[3]{8-x}$$ then you term is given by $$\frac{2\sqrt{a^3-7}-a}{8-a^3}$$ and then multiply numerator and denominator by $$2\sqrt{a^3-7}+a$$  and then you will get
$$\frac{(a-2)(14+7a+4a^2)}{-(a-2)(a^2+2a+4)(2\sqrt{a^3-7}+a)}$$
A: 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:
$$\lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}=\\\lim_{x\to 0}\frac{(2\sqrt{1-x}-2)+(2-\sqrt[3]{8-x})}{x}=\\
\lim_{x\to 0}\frac{2(\sqrt{1-x}-1)}{x}+\lim_{x\to 0}\frac{2-\sqrt[3]{8-x}}{x}=\\
\lim_{x\to 0}\frac{2(1-x-1)}{x(\sqrt{1-x}+1)}+\lim_{x\to 0}\frac{8-(8-x)}{x(4+2\sqrt[3]{8-x}+\sqrt[3]{(8-x)^2})}=\\
-1+\frac1{12}=-\frac{11}{12}.$$
A: In case Taylor series are allowed, then by Taylor expansion around $x=0$:
$$
2\sqrt{1-x} = 2\left(1 - {x\over 2} + O(x^2)\right) \sim 2 - x\\
\sqrt[3]{8-x} = 2 - {x\over 12} + O(x^2)\sim2-{x\over 12}
$$
Thus your limit becomes:
$$
\lim_{x\to0} \frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x} \sim \lim_{x\to0} \frac{2-x - (2-{x\over 12})}{x} = \lim_{x\to 0}\frac{-x + {x\over 12}}{x} = -{11\over 12}
$$
A: Just use Cauchy's rule...
$$
\lim_{x \to 0} \dfrac{2\sqrt{1-x}-\sqrt[3]{8-x}}{x}=
\lim_{x\to 0} \left(-\frac{1}{\sqrt{1-x}}+\frac{1}{3(8-x)^{2/3}} \right)= -1 + \frac{1}{12}=-\frac{11}{12}
$$
A: 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 $-\frac{11}{12}$.
A: $f(x)=2(1-x)^{1/2}$;
$g(x)=\sqrt[3]{8-x}$.
$F(x)=\dfrac{f(x)-f(0)}{x}=$
$\dfrac{2\sqrt{1-x}-2}{x}$
$G(x)=\dfrac{g(x)-g(0)}{x}=\dfrac{\sqrt[3]{8-x}-2}{x}$
$F'(0)= [-(1-x)^{-1/2}]_{x=0}=-1$.
$G'(0)=[-(1/3)(8-x)^{-2/3}]_{x=0}=-(1/3)(1/4)=-1/(12).$
Hence : $F'(0)-G'(0)= -11/(12)$.
Recall:
$\lim_{ x \rightarrow 0}\dfrac{F(x)-F(0)}{x}=F'(0)$, 
similarly for $G$.
A: 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. 
