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Evaluate a limit

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate [itex]\lim_{x \rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}[/itex]

    2. Relevant equations
    The limit laws and the basic rules of algebra.

    3. The attempt at a solution
    I'm completely stuck right at the beginning. I know I have to find a way to get the denominator to be not-zero, but, well, there doesn't seem to be a way to do that (of course there, I just don't see it). Does anyone have a tip to get me started?
     
  2. jcsd
  3. Jun 6, 2012 #2
    Hi, are you familiar with l'Hôpital's rule? In this case, you can derive both the enumerator and the denominator, and then take the limit. If this was new to you, check it out on Wikipedia, for example, it's neat!

    Regards,
    Pifagor
     
  4. Jun 6, 2012 #3
    Hm, let me see if I get this.

    When x goes to 1, both the numerator and the denominator go to zero, therefore, l'Hopital's rule applies if numerator-prime and denominator-prime exist. Deriving the numerator and demoninator ultimately gives [itex]\frac{2\sqrt{x}}{3\sqrt[3]{x^2}}[/itex]. (Right?)

    Intuitively, I can see the limit evaluates to [itex]\frac{2}{3}[/itex] as [itex]x \rightarrow 1[/itex], but I don't quite see how I can algebraically justify turning those different roots in the numerator and denominator into a [itex]\frac{1}{1}[/itex].
     
  5. Jun 6, 2012 #4
    You can solve this without the L'Hospital method too. Try factorizing numerator using [itex]a^2-b^2[/itex] and denominator using [itex]a^3-b^3[/itex] expansions.
     
    Last edited: Jun 6, 2012
  6. Jun 6, 2012 #5
    Ah, I see. I tried that method quickly, and ended up with the same numerator over the sixth root of x times the third root of x minus one... So, yeah. :wink:
     
  7. Jun 6, 2012 #6
    Edited : It works.
     
  8. Jun 6, 2012 #7
    Intuitively, I can see the limit evaluates to [itex]\frac{2}{3}[/itex] as [itex]x \rightarrow 1[/itex], but I don't quite see how I can algebraically justify turning those different roots in the numerator and denominator into a [itex]\frac{1}{1}[/itex].[/QUOTE]

    The thing is, you now have a continuous function in both parts, so the quotient is continuous because the the limit of the denominator is nonzero. Insert x=1, and voilá!

    Pifagor
     
  9. Jun 6, 2012 #8

    Curious3141

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    Homework Helper

    Substitute x = y6, then factorise both numerator and denominator. Find the limit of the expression as y tends to 1.
     
  10. Jun 6, 2012 #9
    That's a great method, and - seeing as how my book had just explained how introducing a new variable can sometimes make something easier to solve - probably exactly what I'm looking for. Thanks. :smile:
     
  11. Jun 6, 2012 #10
    *kicks himself* You're right. I totally forgot just substituting x with 1, instead assuming x could be anything. Oops. :redface:
     
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