Find limit (type 0/0)

Question

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.

Answer 1

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 $$\frac{(a-2)(14+7a+4a^2)}{-(a-2)(a^2+2a+4)(2\sqrt{a^3-7}+a)}$$

Answer 2

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: $$\lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}=\\\lim_{x\to 0}\frac{(2\sqrt{1-x}-2)+(2-\sqrt[3]{8-x})}{x}=\\ \lim_{x\to 0}\frac{2(\sqrt{1-x}-1)}{x}+\lim_{x\to 0}\frac{2-\sqrt[3]{8-x}}{x}=\\ \lim_{x\to 0}\frac{2(1-x-1)}{x(\sqrt{1-x}+1)}+\lim_{x\to 0}\frac{8-(8-x)}{x(4+2\sqrt[3]{8-x}+\sqrt[3]{(8-x)^2})}=\\ -1+\frac1{12}=-\frac{11}{12}.$$

Answer 3

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: $$ \lim_{x\to0} \frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x} \sim \lim_{x\to0} \frac{2-x - (2-{x\over 12})}{x} = \lim_{x\to 0}\frac{-x + {x\over 12}}{x} = -{11\over 12} $$

Answer 4

Just use Cauchy's rule...

$$ \lim_{x \to 0} \dfrac{2\sqrt{1-x}-\sqrt[3]{8-x}}{x}= \lim_{x\to 0} \left(-\frac{1}{\sqrt{1-x}}+\frac{1}{3(8-x)^{2/3}} \right)= -1 + \frac{1}{12}=-\frac{11}{12} $$

Answer 5

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 $-\frac{11}{12}$.

Answer 6

$f(x)=2(1-x)^{1/2}$;

$g(x)=\sqrt[3]{8-x}$.

$F(x)=\dfrac{f(x)-f(0)}{x}=$

$\dfrac{2\sqrt{1-x}-2}{x}$

$G(x)=\dfrac{g(x)-g(0)}{x}=\dfrac{\sqrt[3]{8-x}-2}{x}$

$F'(0)= [-(1-x)^{-1/2}]_{x=0}=-1$.

$G'(0)=[-(1/3)(8-x)^{-2/3}]_{x=0}=-(1/3)(1/4)=-1/(12).$

Hence : $F'(0)-G'(0)= -11/(12)$.

Recall:

$\lim_{ x \rightarrow 0}\dfrac{F(x)-F(0)}{x}=F'(0)$,

similarly for $G$.

Answer 7

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.