How is this limit being solved? I can't grasp it I am going over limits for my finals as I notice this example in my schoolbook discribing limits of the undefined form $0\over0$ in the shape of an irrational fraction. 
$$\lim\limits_{x \to 1} {\sqrt[3]{x}-1\over\sqrt{2x-1}-1} = {0\over0} = \lim\limits_{x \to 1} {(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)(\sqrt{2x-1}+1)\over(\sqrt{2x-1}-1)(\sqrt{2x-1}+1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}$$
I really don't get what is going on here. I know that you are supposed to multiply both the nominator and the denominator by the added value of either. But it appears that here it has been mutliplied by both? And where does this term $(\sqrt[3]{x^2}+\sqrt[3]{x}+1)$ come from?
There is no worked out example, except for the solution after this which is $1\over3$ but thats it.
Could anyone please explain this to me? I would be really gratefull!
Also sorry if I didn't use the correct terms here and there, not a native english speaker.
Thanks in advance.
Cheers, Michiel

$${\sqrt[3]{x}-1\over\sqrt{2x-1}-1} = \frac{\sqrt[3]{x}-1}{\sqrt{2x-1}-1}\cdot \frac{(\sqrt{2x-1}+1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{(\sqrt{2x-1}+1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$
$$=\frac{\left((\sqrt[3]{x}-1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)\right)(\sqrt{2x-1}+1)}{\left((\sqrt{2x-1})^2-1^2\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$
$$=\frac{(x-1)(\sqrt{2x-1}+1)}{2(x-1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}=\frac{\sqrt{2x-1}+1}{2\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$
$$\stackrel{x\to 1}\to \frac{\sqrt{2(1)-1}+1}{2\left(\sqrt[3]{1^2}+\sqrt[3]{1}+1\right)}=\frac{1}{3}$$
