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I currently have $$\lim_{h\to 0}\frac{\sqrt[3]{h+1}-1}{h}$$

Now, I know that when you have square root instead of a cubic root it's easy. You just multiply by the conjugate and simplify afterwards. If it were a sqrt I know that the limit is 1/2. I know that the result here is 1/3 but I can't seem to get there. I always end up with 1/4 due to getting something like: $$\frac{h+1-1}{h(\sqrt[3]{h+1}+1)(\sqrt[3]{h+1}+1)}$$

After simplifying and evaluating the limit you end up with $$\frac{1}{(\sqrt[3]{0+1}+1)(\sqrt[3]{0+1}+1)}$$

which turns out to be $1/4$.

What am I doing wrong?

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Please do check whether my partial edit of your post is correct, and if it is then either re-edit it to LaTeX or ask me to do it for you – DonAntonio Jun 4 '12 at 2:36
You can't conjugate like that with cubic roots. Your algebra is wrong. – Arturo Magidin Jun 4 '12 at 2:38
up vote 6 down vote accepted

You can't conjugate cubic roots the same way you do square roots. You don't get the answer you think you get if you multiply $(\sqrt[3]{h+1}-1)(\sqrt[3]{h+1}+1)(\sqrt[3]{h+1}+1)$. Multiply it out and see that you don't get $h+1-1$: $$\begin{align*} \left(\sqrt[3]{h+1}-1\right)\left(\sqrt[3]{h+1}+1\right)\left(\sqrt[3]{h+1}+1\right) &= \left( \sqrt[3]{h+1}^2 - 1\right)\left(\sqrt[3]{h+1}+1\right)\\ &= \sqrt[3]{h+1}^3 + \sqrt[3]{h+1}^2 - \sqrt[3]{h+1} - 1\\ &= h+1 +(h+1)^{2/3} - (h+1)^{1/3} - 1. \end{align*}$$ So you don't get the concellation you think you get.

If you don't know about derivatives yet, you can do a similar trick to the one used for square roots. When dealing with square roots, you are making use of the identity $$(a+b)(a-b) = a^2-b^2.$$ Here, you want to get rid of a cubic root, so you should make use of the identity $$(a-b)(a^2+ab+b^2) = a^3-b^3.$$ So what we want to do is multiply the numerator and denominator by the factor $$\sqrt[3]{h+1}^2 + \sqrt[3]{h+1} + 1$$ (taking $a=\sqrt[3]{h+1}$ and $b=1$) to get the right cancellation. If we do, we get: $$\begin{align*} \lim_{h\to 0}\frac{\sqrt[3]{h+1} - 1}{h} &= \lim_{h\to 0}\frac{((h+1)^{1/3}-1)((h+1)^{2/3} + (h+1)^{1/3}+1)}{h((h+1)^{2/3} + (h+1)^{1/3} + 1)}\\ &= \lim_{h\to 0}\frac{h+1-1}{h((h+1)^{2/3} + (h+1)^{1/3}+1)}\\ &= \lim_{h\to 0}\frac{h}{h((h+1)^{2/3} + (h+1)^{1/3} + 1)}\\ &= \lim_{h\to 0}\frac{1}{(h+1)^{2/3} + (h+1)^{1/3} + 1}\\ &= \frac{1}{\sqrt[3]{1^2} + \sqrt[3]{1} + 1}\\ &= \frac{1}{1+1+1}\\ &= \frac{1}{3}. \end{align*}$$

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Oh jeesus christ, you are completely right! You don't get the same result if you multiply three times to take the cubic root out. Also, I've seen the code you use to draw the equations on stackexchange. That's great. Anyway. THANKS A TON FOR THE HELP!! – Damieh Jun 4 '12 at 2:53
I'm really curios about the derivative thing though. What do you mean "if you don't know about derivatives yet"? I know a little about them but I'm not really profficient =(. – Damieh Jun 4 '12 at 3:16
@Gaspa79: I'm referring to Chandrasekhar's answer: if you know enough about derivatives to recognize your limit as the derivative of $f(x) = \sqrt[3]{x+1}$ at $x=1$, then you can compute the limit by computing the derivative. But if you don't know enough to recognize this limit as the derivative, and to be able to compute the derivative without the limit in the first place... – Arturo Magidin Jun 4 '12 at 3:18

The problem is that $(\sqrt[3]{h+1}-1)(\sqrt[3]{h+1}+1)^2$ is not the factorization of $(h+1)-1$, which is the end result you desire. Rather, for cubes, $u^3-1=(u-1)(u^2+u+1)$ is the correct "level higher" version of $u^2-1=(u-1)(u+1)$. Indeed, in general we have $u^n-1=(u-1)(u^{n-1}+\cdots+1)$, which is essentially the geometric sum formula. Therefore you want to evaluate


at $h=0$ (note we have $u=\sqrt[3]{h+1}$ here), which I have faith you can do. :-)

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Thanks a ton for the help man! – Damieh Jun 4 '12 at 2:53

I think you have multiplied incorrectly. I suppose you want to find $$\lim_{h \to 0 } \frac{(h+1)^{1/3} -1}{h}$$

Note that this is nothing but the derivative of the function $f(x) = (x+1)^{1/3}$ at $x =0$. So your $f'(x) = \frac{1}{3} \cdot (x+1)^{-2/3}$. Now it's easy to see that $f'(0)=\frac{1}{3}$ and so your limit is $\frac{1}{3}$.

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Another way without knowing before hand the derivative:$$A^3-1=(A-1)(A^2+A+1)\Longrightarrow A-1=\frac{A^3-1}{A^2+A+1}$$and putting $\,A=\sqrt[3]{h+1}\,$ one gets:$$\frac{\sqrt[3]{h+1}-1}{h}=\frac{\rlap{/}{h}+1-1}{\rlap{/}{h}\left(\sqrt[3]{(h+1)^2}+\sqrt[3]{h+1}+1\right)}\to\frac{1}{1+1+1}=\frac{1}{3}$$

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