# Limit without using L'Hopital's rule.

1. Dec 20, 2012

### cambo86

1. The problem statement, all variables and given/known data
$$\lim_{x\rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}$$

I can do this very easily using L'Hopital's rule but in the text book I'm going through it is a problem given before L'Hopital's rule is taught. Is there a way of doing this without using L'Hopita'ls rule?

Last edited: Dec 20, 2012
2. Dec 20, 2012

### Curious3141

One way is to let $x = (y+1)^2$, after which the numerator can be simplified with the generalised binomial theorem.

3. Dec 20, 2012

### SammyS

Staff Emeritus
Do you know how to factor the difference of squares and the difference cubes ?

$(a-b)(a+b)=a^2-b^2$

$(c-d)(c^2+cd+d^2)=c^3-d^3$

Therefore, $\ \ (\sqrt{x\,}-1)(\sqrt{x\,}+1)=x-1$

and $\ \ (\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)=x-1\ .$

These can be used to cancel x-1 in the numerator & denominator.

4. Dec 20, 2012

### cambo86

Thanks, both responses were very helpful.