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I stumbled upon a limit some time ago today, which I've tried solving with no success: $$\lim_{x \to \infty} f(x) = \frac{\sqrt[3]{x^3+2}+\sqrt[3]{8x^3+1}-3x}{2}=0$$ Which presents an $\infty-\infty$ indeterminate form. The main idea that I had, and that I definitely thought would solve the problem, was factorizing the cubic roots (yes those are cubic roots, sorry if it can't be seen properly).

Basically, you know $a^3+b^3=(a+b)(a^2-ab+b^2)$.

Which, taking cubic roots, gives: $a+b=(\sqrt[3]{a}+\sqrt[3]{b})(\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2})$.

So, you can clear that and obtain: $$\sqrt[3]{a}+\sqrt[3]{b}=\frac{a+b}{\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}}$$ I tried that method both by doing the addition of both cubic roots, and by doing the substraction of one cubic root and the other term (putting it as $\sqrt[3]{27x^3}$), only to find out yet another indetermination, so all that was to no avail.

Does anyone have any idea on how to solve that limit, or prove it is actually $0$ (which I got by seeing the plot of the function)?

Thanks in advance.

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    $\begingroup$ Please divide each term in the numerator by $x^3$ under the cube root sign and by x outside it. And then multiply thru by x. All will become clear. $\endgroup$ – Betty Mock Dec 12 '13 at 2:05
  • $\begingroup$ @BettyMock thanks for the tip, I used it when solving the limit and it made things much easier as you said ;) $\endgroup$ – F.Webber Dec 12 '13 at 2:34
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Hint: The top is $$(\sqrt[3]{x^3+2}-x) +(\sqrt[3]{8x^3+1}-2x).$$ Now the identity you quote (negative version), applied to the parts, should lead to success.

Another way: Or else we can use L'Hospital's Rule. This is not a joke. Divide top and bottom by $x$, and make the substitution $t=1/x$. We get $$\frac{\sqrt[3]{1+2t^3}+\sqrt[3]{8+t^3}-3}{2t}.$$ Now L'Hospital's Rule works nicely. Or the Taylor series expansion.

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  • $\begingroup$ This is the best method among the answers +1! $\endgroup$ – Christopher K Dec 12 '13 at 2:09
  • $\begingroup$ Oh very clever movement there, I used your first suggestion, complemented with the others advice of multiplying and dividing by x, and it turned out to work great! Thanks again, you've helped me twice today :D $\endgroup$ – F.Webber Dec 12 '13 at 2:32
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    $\begingroup$ You are welcome. For the first way, there is no need to multiply, divide by $x$. For example, with the $\sqrt[3]{x^3+2}-x$ part, when you multiply top and bottom by "$a^2+ab+b^2$" ($a=\sqrt[3]{x^3+2}$, $b=x$ you end up with $\frac{2}{a^2+ab+b^2}$, and the bottom blows up. $\endgroup$ – André Nicolas Dec 12 '13 at 2:39
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Hint The problem is $$a^3+b^3=(a+b)(a^2-ab+b^2)$$

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  • $\begingroup$ Whoops sorry, I actually copied it wrong here, but I had indeed used that expression. Sorry for the confusion :S $\endgroup$ – F.Webber Dec 12 '13 at 2:07
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?use the binomial theorem: $$ (1 + z)^{\frac13} = 1 + \frac13z + ... $$

then use $$ \sqrt[3]{x^3+2} = x\sqrt[3]{1+\frac2{x^3}} $$ and similarly for the other surd

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  • $\begingroup$ Thanks man :D Upvoted because that second tip was crucial to be able to simplify it properly. $\endgroup$ – F.Webber Dec 12 '13 at 2:33
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    $\begingroup$ welcome Elucas. like politics (ONLY in this respect) math is the art of the possible. many questions will maybe never be answerable, but the realm of the soluble forever expands. here you just needed the right nutcracker for the job. i can feel how exciting it must have been in the 17th century to suddenly be able to use the binomial theorem for indices outside $\mathbb{N}$, though i suspect the influence in those times of knowledge and techniques transmitted from the Kerala school in S India (via Jesuit missionaries?) remains very underestimated $\endgroup$ – David Holden Dec 12 '13 at 8:57

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