Finding limit of cube root 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.

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: $$(a-b)(a^2+ab+b^2)=a^3-b^3$$
