Using conjugates to find a limit with a cubic root: $\lim_{h\to 0}\frac{\sqrt[3]{h+1}-1}{h}$

Question

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?

Answer 1

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: $$\begin{align*} \lim_{h\to 0}\frac{\sqrt[3]{h+1} - 1}{h} &= \lim_{h\to 0}\frac{((h+1)^{1/3}-1)((h+1)^{2/3} + (h+1)^{1/3}+1)}{h((h+1)^{2/3} + (h+1)^{1/3} + 1)}\\ &= \lim_{h\to 0}\frac{h+1-1}{h((h+1)^{2/3} + (h+1)^{1/3}+1)}\\ &= \lim_{h\to 0}\frac{h}{h((h+1)^{2/3} + (h+1)^{1/3} + 1)}\\ &= \lim_{h\to 0}\frac{1}{(h+1)^{2/3} + (h+1)^{1/3} + 1}\\ &= \frac{1}{\sqrt[3]{1^2} + \sqrt[3]{1} + 1}\\ &= \frac{1}{1+1+1}\\ &= \frac{1}{3}. \end{align*}$$

Answer 2

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 $\frac{1}{3}$.

Answer 3

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

$$\frac{\sqrt[3]{h+1}-1}{h}\frac{(\sqrt[3]{h+1})^2+\sqrt[3]{h+1}+1}{(\sqrt[3]{h+1})^2+\sqrt[3]{h+1}+1}=\frac{(\sqrt[3]{h+1})^3-1}{h\big((\sqrt[3]{h+1})^2+\sqrt[3]{h+1}+1\big)}$$

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

Answer 4

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:$$\frac{\sqrt[3]{h+1}-1}{h}=\frac{\rlap{/}{h}+1-1}{\rlap{/}{h}\left(\sqrt[3]{(h+1)^2}+\sqrt[3]{h+1}+1\right)}\to\frac{1}{1+1+1}=\frac{1}{3}$$