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How can I calculate this limit? $$\lim_\limits{x\to2}\frac{\sqrt{12-x^3}-\sqrt[3]{x^2+4}}{x^2-4}$$

Context

I should do it without using L'Hospital's rule. Factoring $a^n - b^n=(a-b)(\dots)$ doesn't seem to help in this example: the numerator contains two different powers, $1/2$ and $1/3$..

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  • $\begingroup$ L'Hospital's Rule ... $\endgroup$ – Angelo Jan 26 '15 at 19:49
  • $\begingroup$ Sorry but without this $\endgroup$ – GorillaApe Jan 26 '15 at 19:50
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    $\begingroup$ Are you sure the first term in the numerator isn't a cube root? $\endgroup$ – Tim Raczkowski Jan 26 '15 at 19:56
  • $\begingroup$ yes I am sure... $\endgroup$ – GorillaApe Jan 26 '15 at 20:00
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    $\begingroup$ Write $x=2+t$ with $t=o(1)$ then $$x^2-4\sim4t,$$ and $12-x^3=4(1-3t+o(t))$ hence $$\sqrt{12-x^3}=2\sqrt{1-3t+o(t)}=2-3t+o(t),$$ and $x^2+4=8(1+\frac12t+o(t))$ hence $$\sqrt[3]{x^2+4}=2+\tfrac13t+o(t),$$ from which the limit $(-3-\frac13)\cdot\frac14$ is apparent. $\endgroup$ – Did Jan 26 '15 at 20:04
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$$\lim_\limits{x\to2}\frac{\sqrt{12-x^3}-\sqrt[3]{x^2+4}}{x^2-4}= \\ =\lim_\limits{x\to2}\frac{\left(\sqrt{12-x^3}-2\right)+\left(2-\sqrt[3]{x^2+4}\right)}{x^2-4} = \\ =\lim_\limits{x\to2}\frac{\frac{\left(\sqrt{12-x^3}-2\right)\left(\sqrt{12-x^3}+2\right)}{\sqrt{12-x^3}+2} + \frac{\left(2-\sqrt[3]{x^2+4}\right)\left(4+2\sqrt[3]{x^2+4} + \left(\sqrt[3]{x^2+4} \right)^2\right)}{4+2\sqrt[3]{x^2+4} + \left(\sqrt[3]{x^2+4} \right)^2}}{x^2-4}= \\ =\lim_\limits{x\to2}\frac{8-x^3}{\left(\sqrt{12-x^3}+2\right)(x-2)(x+2)}\\+ \lim_\limits{x\to2}\frac{4-x^2}{(x^2-4)\left( 4+2\sqrt[3]{x^2+4} + \left(\sqrt[3]{x^2+4} \right)^2\right)} =\ldots$$ Can you continue?

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  • $\begingroup$ very smart... wow $\endgroup$ – GorillaApe Jan 26 '15 at 20:23
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Let's test that theory. Let $a=\sqrt{12-x^3}$ and $b=\sqrt[3]{x^2+4}$. Now let's multiply top and bottom by $a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$. So our new numerator is

$$a^6-b^6=(12-x^3)^3-(x^2+4)^2=1728-432x^3+36x^6-x^9-x^4-8x^2-16=$$ $$-x^9+36x^6-x^4-432x^3-8x^2+1712$$

It's an unenviable task, but Wolfram verifies this is a multiple of $x-2$, so you can divide out this factor. The new denominator evaluated at $2$ will be $(2+2)(6\times2^5)=768$. The messy part is doing the polynomial division, then evaluating a degree $8$ polynomial at $2$...

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    $\begingroup$ seems possible but hard $\endgroup$ – GorillaApe Jan 26 '15 at 20:37
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Make the change of variable $x\to x+2$.

$$\frac{\sqrt{4-x^3-6x^2-12x}-\sqrt[3]{x^2+4x+8}}{x^2+4x}.$$

Using Taylor up to degree $1$ suffices

$$\frac{2\sqrt{1-3x}-2\sqrt[3]{1+\frac12x}}{4x}=\frac{2(1-\frac{3}{2}x)-2(1+\frac{1}{6}x)}{4x}=-\frac56.$$

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  • $\begingroup$ Isn't it $-5/6$? $\endgroup$ – Angelo Jan 27 '15 at 12:14
  • $\begingroup$ @Angelo: yep, I had a sign error at the denominator. Thanks. $\endgroup$ – Yves Daoust Jan 27 '15 at 13:19

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