<s>[INST] 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?
 [/INST] Another way : change variable $x=1+y$; so $$A=\frac{\sqrt x -  1}{ \sqrt[3] x -  1}=\frac{\sqrt{1+y} -  1}{ \sqrt[3] {1+y} -  1}$$ Now, using the fact that, close to $y=0$ (using the generalized binomial theorem as lulu commented)  $$(1+y)^a=1+a y+\frac{1}{2}a \left(a-1\right) y^2+O\left(y^3\right)$$  which makes $$A=\frac{1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)-1}{1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)-1}\approx \frac{\frac{y}{2}-\frac{y^2}{8}}{\frac{y}{3}-\frac{y^2}{9}}=\frac{\frac{1}{2}-\frac{y}{8}}{\frac{1}{3}-\frac{y}{9}}$$ Now make $y\to0$ to get the result.
You can even get more if you know long division. Omitting the high order terms, the last expression is $\sim\frac{3}{2}+\frac{y}{8}$ which reveals not only the limit but also how it is approached.
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