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I'm trying to evaluate this limit, but I don't think it's coming out correctly. Could someone please offer me some assistance?

Evaluate limit analytically $$\lim_{h\to 0}\frac{\sqrt[3]{x + h} - \sqrt[3]{x}}{h}.$$

What I did was multiply $(x+h)^{2/3} + x^{2/3}$ top and bottom to get

$$\lim_{h\to 0}\frac{(x+h)-x}{h((x+h)^{2/3} + x^{2/3})}.$$

I end up getting $\dfrac{1}{2x^{2/3}}$.

The reason why I don't think I did this write is because isn't the limit above the definition of a derivative? And if so, then isn't the derivative of $\sqrt[3]{x}$ equal to $\dfrac{1}{3x^{2/3}}$?

I would really appreciate any kind of help. Thanks.

Michael Albanese
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Astro
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    Instead, multiply top and bottom by $(x+h)^{2/3}+(x+h)^{1/3}x^{1/3}+x^{2/3}$. This takes advantage of the identity $s^3-t^3=(s-t)(s^2+st+t^2)$. – André Nicolas Jun 19 '14 at 23:05
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    Note that with your choice we do not get $(x+h)-x$ on top, algebra glitch. – André Nicolas Jun 19 '14 at 23:11
  • Thank you for pointing out my mistake! I appreciate it. Also, thank you for the link to the other website. Definitely was the same question :) – Astro Jun 19 '14 at 23:27

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Problem is, you didn't FOIL it out all the way in the numerator: $$\left(\sqrt[3]{x + h} - \sqrt[3]{x}\right)\left((x+h)^{2/3} + x^{2/3}\right)=(x+h)+\underbrace{\sqrt[3]{x+h}\,x^{2/3}-(x+h)^{2/3}\sqrt[3]{x}}_{\textit{does not cancel}}-x$$

You're trying to use a difference of squares formula here, something like $(a-b)(a+b)$ with $a=\sqrt[3]{x+h}$ and $b=\sqrt[3]{x}$. But you don't want to square $\sqrt[3]{x}$, you want to cube it! So, instead, use a difference of cubes formula: $$(a-b)(a^2+ab+b^2)=a^3-b^3$$

Alexander Gruber
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