How can I solve the folowing limit using only algebric manipulation:
limit when x aproches 0 of: {[x^(1/3)]-1}/{x^(1/2)]-1}
Question: Why do you say this is difficult?
Do you realize that $\displaystyle \displaystyle \lim _{x \to 0} \sqrt x $ does not exist because $\displaystyle \sqrt{x}$ is not defined for $\displaystyle x<0$.
But $\displaystyle \displaystyle \lim _{x \to 0^+} \sqrt x=0 $.
So you may want to double check the statement of this question.
Let's make the problem a bit more interesting and assume that the limit is supposed to be as x goes to 1 rather than 0. You can then check using l'Hôpital's rule that the answer should be 2/3. But it has to be done using only algebraic manipulation, so l'Hôp is not allowed.
Let $\displaystyle u = x^{1/6}$. Then as x goes to 1, u also goes to 1. So we want to find
$\displaystyle \displaystyle \lim_{x\to 1}\frac{x^{1/3}-1}{x^{1/2}-1} = \lim_{u\to 1}\frac{u^2-1}{u^3-1} = \lim_{u\to 1}\frac{(u-1)(u+1)}{(u-1)(u^2+u+1)} = \lim_{u\to 1}\frac{u+1}{u^2+u+1} = \frac23.$
Edit. Beaten by Danny.