Evaluating $\lim _{x\to 0}\frac{\left(x^3-x\right)}{\sqrt[3]{x^2-8}-x+2}$ without L'Hospital's Theorem?

Question

Couldn't solve it without L'Hospital's Theorem, need help with any alternative solution.

$\lim _{x\to 0}\frac{\left(x^3-x\right)}{\sqrt[3]{x^2-8}-x+2}$

Answer 1

Rationalizing the denominator:

$$ \begin{aligned} \lim_{x\to 0}\frac{\left(x^3-x\right)}{\sqrt[3]{x^2-8}-x+2}&=\lim_{x\to 0} \frac{x^3-x}{\frac{x^2-8-(x-2)^3}{\sqrt{(x^2-8)^2}+(x-2)\sqrt[3]{x^2-8}+(x-2)^2}}\\ &= \lim_{x\to 0} \frac{(x^3-x)(\sqrt{(x^2-8)^2}+(x-2)\sqrt[3]{x^2-8}+(x-2)^2)}{-x(x^2-7x+12)}\\ &=\lim_{x\to 0} \frac{(x^2-1)(\sqrt[3]{(x^2-8)^2}+(x-2)\sqrt[3]{x^2-8}+(x-2)^2)}{-(x^2-7x+12)}\\ &=1 \end{aligned} $$

Edit: We can also do it with l'Hopital:

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

just to check this is the right answer.

Answer 2

Using $a^3-b^3=(a-b)(a^2+ab+b^2) $, \begin{align} \frac{\left(x^3-x\right)}{\sqrt[3]{x^2-8}-x+2} &=\frac{\left(x^3-x\right)}{\sqrt[3]{x^2-8}-x+2}\frac {(x^2-8)^{2/3}+(x^2-8)^{1/3}(x-2)+(x-2)^{2}}{(x^2-8)^{2/3}+(x^2-8)^{1/3}(x-2)+(x-2)^{2}}\\ \ \\ &=\frac{\left(x^3-x\right)( (x^2-8)^{2/3}+(x^2-8)^{1/3}(x-2)+(x-2)^{2})} {x^2-8-(x-2)^3}\\ \ \\ &=\frac{\left(x^3-x\right)( (x^2-8)^{2/3}+(x^2-8)^{1/3}(x-2)+(x-2)^{2})} {-x^3+7x^2-12x}\\ \ \\ &=\frac{\left(x^2-1\right)( (x^2-8)^{2/3}+(x^2-8)^{1/3}(x-2)+(x-2)^{2})} {-x^2+7x-12}\\ \ \\ &\to\frac {(-1)(4+(-2)(-2)+4)}{-12}=1. \end{align}

Answer 3

You can first consider the reciprocal: $$\frac{\sqrt[3]{x^2-8}+2-x}{x(x^2-1)}=\frac 1{x^2-1}\left(\frac{\sqrt[3]{x^2-8}+2}{x}-1\right)$$ so, the only critical term for $x\to 0$ is $\frac{\sqrt[3]{x^2-8}+2}{x}$.

With $t^3=x^2-8$ and, hence, $t \to -2^+$ you get $x=\pm\sqrt{t^3+8}$ and $$\frac{\sqrt[3]{x^2-8}+2}{x} = \frac{t+2}{\pm\sqrt{t^3+8}}=\ldots$$ Now use $t^3+8=(t+2)(t^2-2t+4)$:

$$\ldots =\frac{\pm\sqrt{t+2}}{\sqrt{t^2-2t+4}}\stackrel{t\to-2^+}{\longrightarrow}0$$

So, the limit of the critical term is $0$ and, hence, the limit searched for is the reciprocal of $$\frac 1{0-1}\left(0-1\right)=1$$

so, the limit is $\boxed{1}$.

Answer 4

$\quad$Acutally, you can use Taylor formula to obtain this limit.

$\lim_{x \to 0} \frac{x^3-x}{\sqrt[3]{x^2-8} -(x-2)}$ =$\lim_{x \to 0} \frac{x(x+1)(x-1)}{\sqrt[3]{x^2-8} -(x-2)}$=$\lim_{x \to 0} \frac{-x}{\sqrt[3]{x^2-8} -(x-2)}$

=$\lim_{x \to 0} \frac{1}{1-\frac{2+\sqrt[3]{x^2-8}}{x}}$

$\quad$According to the equation above, we can determine the result if we could prove the result that the limit$\lim_{x \to 0}\frac{2+\sqrt[3]{x^2-8}}{x} = 0$. Hence we will use the Taylor formula to get this conculusion.

If both the numerator and denominator are divided by $-2$, then

$\lim_{x \to 0}\frac{2+\sqrt[3]{x^2-8}}{x} $=$\lim_{x \to 0}\frac{-1+\sqrt[3]{\frac{-x^2}{8}+1}}{\frac{-x}{2}} = 0$, which is equivalent to $\lim_{x \to 0} \frac{-1+1+\frac{-x^2}{8}+o(x^2)}{\frac{-x}{2}}=0$. Therefore, we get what we desire.

I am so sorry for my bad style of typing. I'm not so familiar with LaTeX , but I hope You enjoy my solution.

Answer 5

Assuming that you would get more than the limit itself, consider $$y=\frac{x^3-x}{\sqrt[3]{x^2-8}-x+2}$$ and work the pieces using the binomial expansion $$\sqrt[3]{x^2-8}=-2+\frac{x^2}{12}+\frac{x^4}{288}+O\left(x^6\right)$$ $${\sqrt[3]{x^2-8}-x+2}=-x+\frac{x^2}{12}+\frac{x^4}{288}+O\left(x^6\right)$$ $$y=\frac{x^3-x}{-x+\frac{x^2}{12}+\frac{x^4}{288}+O\left(x^6\right)}$$ Now, the long division to get $$y=1+\frac{x}{12}-\frac{143 x^2}{144}+O\left(x^3\right)$$

Try it for $x=\frac 12$. The exact value is $0.783$ while the above expansion gives $0.793$.