How do I simplify and evaluate the limit of $(\sqrt x - 1)/(\sqrt[3] x - 1)$ as $x\to 1$? Consider this limit:
$$ \lim_{x \to 1} \frac{\sqrt x -  1}{ \sqrt[3] x -  1} 
$$ 
The answer is given to be 2 in the textbook. 
Our math professor skipped this question telling us it is not in our syllabus, but how can it be solved?

Note the following identities,
$$ y^2-1 = (y-1)(y+1) $$
$$ y^3-1 = (y-1)(y^2+y+1) $$
we can use these to rewrite the numerator and the denomiantor by substituting $\sqrt{x}$ and $\sqrt[3]{x}$ for $y$ respectively,
$$ x - 1 = (\sqrt{x}-1)(\sqrt{x}+1) \Rightarrow \sqrt{x}-1 = \frac{x-1}{\sqrt{x}+1} $$
$$ x - 1 = (\sqrt[3]{x}-1)(\sqrt[3]{x}^2+\sqrt[3]{x}+1) \Rightarrow \sqrt[3]{x}-1 = \frac{x-1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1} $$
substituting this into the ratio you're taking the limit of we get,
$$ \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{\sqrt[3]{x}-1} = \lim_{x\rightarrow 1}  \frac{x-1}{\sqrt{x}+1} \frac{\sqrt[3]{x}^2+\sqrt[3]{x}+1} {x-1} = \frac32 $$
