Please help me with this limit $$
\lim_{h\to 0} {\frac {\sqrt[3]{x+h}-\sqrt[3]{x}} h}
$$
I tried with the conjugation in the top and under but i still have the determiantion the result must be 1/3 i did it with Matlab, you CAN'T use Lhopital
 A: By definition of the derivative $\lim_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{h}=f'(x),$
this limit is the derivative of function $x \mapsto f(x)=\sqrt[3]{x}=x^{1/3}$.
No wonder that the result is $\dfrac{1}{3}x^{-2/3}.$
A: Just multiply both side to $ \left( \sqrt [ 3 ]{ { \left( x+h \right)  }^{ 2 } } +\sqrt [ 3 ]{ x\left( x+h \right)  } +\sqrt [ 3 ]{ { x }^{ 2 } }  \right) $ to get announced result $$\lim _{ h\to 0 }{ \frac { \sqrt [ 3 ]{ x+h } -\sqrt [ 3 ]{ x }  }{ h }  } =\lim _{ h\to 0 }{ \frac { \left( \sqrt [ 3 ]{ x+h } -\sqrt [ 3 ]{ x }  \right) \left( \sqrt [ 3 ]{ { \left( x+h \right)  }^{ 2 } } +\sqrt [ 3 ]{ x\left( x+h \right)  } +\sqrt [ 3 ]{ { x }^{ 2 } }  \right)  }{ h\left( \sqrt [ 3 ]{ { \left( x+h \right)  }^{ 2 } } +\sqrt [ 3 ]{ x\left( x+h \right)  } +\sqrt [ 3 ]{ { x }^{ 2 } }  \right)  }  } =\\ =\lim _{ h\to 0 }{ \frac { h }{ h\left( \sqrt [ 3 ]{ { \left( x+h \right)  }^{ 2 } } +\sqrt [ 3 ]{ x\left( x+h \right)  } +\sqrt [ 3 ]{ { x }^{ 2 } }  \right)  }  } =\frac { 1 }{ 3\sqrt [ 3 ]{ { x }^{ 2 } }  } $$
A: I elaborate on the answer by haqnatural to provide a more general insight.
Consider:
$$\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$$
You know that it is solved by conjugation. But call $t=\sqrt{x+h}$, $a=\sqrt{x}$, so $h=t^2-a^2$ and $h\to 0 \Rightarrow t \to a$. Now the limit is written as:
$$\lim_{t\to a} \frac{t-a}{t^2-a^2}$$
and it is all about factorizing polynomials: $t^2-a^2=(t-a)(t+a)$.
In the same spirit, set $t=\sqrt[3]{x+h}$, $a=\sqrt[3]{x}$ and write your limit as:
$$\lim_{h\to 0} \frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}=\lim_{t\to a} \frac{t-a}{t^3-a^3}$$
Now $a$ is a root of the polynomial in the denominator so it can be factorized:
$$t^3-a^3=(t-a) P(t)$$
Find $P(t)$ by division and simplify.
My view is that somehow, whenever you have $\frac{0}{0}$ you have to factorize (sort of) polynomials and simplify the "zeros".
