Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Nov 16, 2008 #1
    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. jcsd
  3. Nov 16, 2008 #2

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Re: limit

    have you done taylor series?
     
  4. Nov 16, 2008 #3
    Re: limit

    Yes but we have to do this by changing the variables.
     
  5. Nov 16, 2008 #4

    HallsofIvy

    User Avatar
    Science Advisor

    Re: limit

    Looks to me like a problem where you would want to "rationalize the denominator.
    [itex]a^3- b^3= (a- b)(a^2+ ab+ b^2)[/itex] so to "rationalize" [itex]^3\sqrt{y^2}- 1[/itex], you need to multiply by [itex]^3\sqrt{y^4}+ ^3\sqrt{y^2}+ 1[/itex]. Multiplying the numerator and denominator by that will make the denominator [itex]y^2- 1[/itex]. You should be able to factor a "y- 1" out of the numerator also to cancel the "y- 1" in the denominator,.
     
    Last edited by a moderator: Nov 16, 2008
  6. Nov 16, 2008 #5
    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?
     
  7. Nov 16, 2008 #6

    HallsofIvy

    User Avatar
    Science Advisor

    Re: limit

    Assuming you the calculations correctly, which you did not show, yes.
     
  8. Nov 17, 2008 #7
    Re: limit

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

    Thank you.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook