1. ## difficult limit

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}

2. Originally Posted by Mppl
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}
Dear Mppl,

Do you mean, $\displaystyle\lim_{x\rightarrow{0}}\frac{x^{\frac{ 1}{3}}-1}{x^{\frac{1}{2}}-1}$ ??

3. yes precisely that one. Can you help me with that?

4. Question: Why do you say this is difficult?
Do you realize that $\displaystyle \lim _{x \to 0} \sqrt x$ does not exist because $\sqrt{x}$ is not defined for $x<0$.
But $\displaystyle \lim _{x \to 0^+} \sqrt x=0$.

So you may want to double check the statement of this question.

5. well lets admit its 0+ then...whats the result then? and how to get there?

6. Originally Posted by Mppl
well lets admit its 0+ then...whats the result then? and how to get there?
$\displaystyle \lim _{x \to 0^+} \sqrt x=0$.
$\displaystyle \lim _{x \to 0^+} \sqrt[3] x=0$.

Now you finish.

7. oh my bad the limit is x aproching 1 and not 0 sorry...what would be the solution there?

8. Let $x = y^6$ so then you have

$\displaystyle \lim_{y \to 1} \dfrac{y^2-1}{y^3-1}$

which you can factor but L'Hopital's rule would be faster.

9. 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 $u = x^{1/6}$. Then as x goes to 1, u also goes to 1. So we want to find

$\displaystyle \lim_{x\to 1}\frac{x^{1/3}-1}{x^{1/2}-1} = \lim_{u\to 1}\frac{u^2-1}{u^3-1} = \lim_{u\to 1}\frac{(u-1)(u+1)}{(u-1)(u^2+u+1)} = \lim_{u\to 1}\frac{u+1}{u^2+u+1} = \frac23.$

Edit. Beaten by Danny.

10. thanks a lot men