1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with limits problem

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data
    lim[tex]\frac{x^{1/3}-a^{1/3}}{x - a}[/tex] as a approaches 0

    2. Relevant equations



    3. The attempt at a solution

    Heres my problem, I can do limits no problem, but my book shows steps on how to do the problem, there is one particular step I dont understand. It is how they got from the above equation to

    lim[tex]\frac{x^{1/3}-a^{1/3}}{(x^{1/3}-a^{1/3})(x^{2/3}+x^{1/3}a^{1/3}+a^{2/3})}[/tex]


    ok so if I multiple the denominator out I get x - a, so it seems like they changed the denominator to a polynomial that they could factor out the same value as the numerator yet still = x - a. Am I understanding this right?

    edit - sorry I butchered the format some due to previewing it my post 23432 times.
     
  2. jcsd
  3. Feb 11, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Limits

    Yes, you are understanding it right. x-a=(x^(1/3))^3-(a^(1/3))^3. They factored x-a to to cancel the numerator.
     
  4. Feb 11, 2010 #3
    Re: Limits

    Ok, but how did they come up with that polynomial? They needed to find a value they could multiply the numerator by to get the value in the denominator? Is there some fundamental thing I should know about how they came up with this polynomial?
     
  5. Feb 11, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Limits

    a^3-b^3=(a-b)*(a^2+ba+b^2). They just expressed x-a as a difference of cubes, which were, of course, the cubes of x^(1/3) and a^(1/3). It's not that fundamental. It's just an algebraic trick.
     
  6. Feb 11, 2010 #5
    Re: Limits

    Great! Thanks Dick.
     
  7. Feb 11, 2010 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Limits

    They just used that (a^3-b^3)=(a-b)*(a^2+ab+b^2). There are similar factorizations for (a^n-b^n) for any n. It's just an algebraic trick.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help with limits problem
Loading...