Question about Limits (something i have not seen before) $$\lim_{x\to 1} \frac{x-1}{\sqrt[3]{x}-1} $$
I know that first step is to rationalize it with $\sqrt[3]{x}-1$ top and bottom
but I don't understand why did the professor add $\sqrt[3]{x^2}$ both top and bottom?
(sorry for not being able to format I will post the picture so it is easier
for you to understand) 
Why did he add that part that I marked with red ?
 A: In order to get rid of the cubic root, you can try multiplying numerator and denominator with a suitable factor so you can use common identities.
To try and make use of the identity $(a-b)(a^2+ab+b^2) = a^3-b^3$, choose $a=\sqrt[3]{x}$ and $b=1$, then the red part is precisely the term $a^2$ from the factor $a^2+ab+b^2$.
A: It comes from the identity $$a^3-b^3 = (a-b)(a^2+ab+b^2),$$with $a = \sqrt[3]{x}$ and $b = 1$. You already have the $a-b$ part (which is $\sqrt[3]{x}-1$), but it is easier to work with $a^3-b^3$. It a more complicated kind of "conjugate". You have: $$\begin{align}\lim_{x \to 1}\frac{x-1}{\sqrt[3]{x}-1} &=\lim_{x \to 1}\frac{(x-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)} \\ & = \lim_{x \to 1}\frac{(x-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{x-1} \\ &= \lim_{x \to 1}(\sqrt[3]{x^2}+\sqrt[3]{x}+1) = 3\end{align}$$
The same goes if you had, for example, $$\lim_{x \to 1}\frac{x-1}{\sqrt[4]{x}-1},$$ you would multiply both numerator and denominator by $$\sqrt[4]{x^3}+\sqrt[4]{x^2}+\sqrt[4]{x}+1,$$ thinking of $$a^4-b^4 = (a-b)(a^3+a^2+a+1),$$with $a = \sqrt[4]{x}$ and $b=1$. And so on.
A: Because professor have used this formula:
$$(a-b)(a^2+a \cdot b+b^2) = a^3 - b^3$$ 
and this task $a = \sqrt[3]{x}$ and $ b = 1 $ 
If you do a little computation I will understand it easily.
A: HINT:
Let $\sqrt[3]x-1=y\implies x=(y+1)^3$
As $x\to1,y\to0$
This method can be employed safely for $\sqrt[n]x-1$
Alternatively, $$\lim_{x\to1}\frac{\sqrt[3]x-1}{x-1}=\dfrac{d(x^{1/3})}{dx}_{(\text{ at }x=1)}$$
