# Find the limit of (sqrt(1+x) - 1)/(cuberoot(1+x)-1) as x->0

1. Nov 16, 2008

### sara_87

Find the limit of (sqrt(1+x) - 1)/(cuberoot(1+x)-1) as x-->0

1. The problem statement, all variables and given/known data

Fin limit of the follwing as x tends to 0:

[sqrt(1+x)-1]/[cuberoot(1+x)-1]

2. Relevant equations

3. The attempt at a solution

using change of variables:
let y=sqrt(1+x)
so = limit (as y tends to 1): [y-1]/cuberoot(y^2)-1]

Then i got stuck.

any help woyld be v much appreciated.
Thank you.

2. Nov 16, 2008

### malawi_glenn

Re: limit

have you done taylor series?

3. Nov 16, 2008

### sara_87

Re: limit

Yes but we have to do this by changing the variables.

4. Nov 16, 2008

### HallsofIvy

Staff Emeritus
Re: limit

Looks to me like a problem where you would want to "rationalize the denominator.
$a^3- b^3= (a- b)(a^2+ ab+ b^2)$ so to "rationalize" $^3\sqrt{y^2}- 1$, you need to multiply by $^3\sqrt{y^4}+ ^3\sqrt{y^2}+ 1$. Multiplying the numerator and denominator by that will make the denominator $y^2- 1$. You should be able to factor a "y- 1" out of the numerator also to cancel the "y- 1" in the denominator,.

Last edited: Nov 16, 2008
5. Nov 16, 2008

### sara_87

Re: limit

Ok so after i cancel the y-1, i get:
(cuberoot(y^2)+1)/(y+1)
so substituting the y=1, i should get limit =1
is that right?

6. Nov 16, 2008

### HallsofIvy

Staff Emeritus
Re: limit

Assuming you the calculations correctly, which you did not show, yes.

7. Nov 17, 2008

### sara_87

Re: limit

since after cancelling the y-1,this will give: (cuberoot(y^2)+1)/(y+1)=2/2=1

Thank you.