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Limit without using L'Hopital's rule.

  1. Dec 20, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]
    \lim_{x\rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}
    [/tex]

    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. jcsd
  3. Dec 20, 2012 #2

    Curious3141

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    One way is to let ##x = (y+1)^2##, after which the numerator can be simplified with the generalised binomial theorem.
     
  4. Dec 20, 2012 #3

    SammyS

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

    [itex](a-b)(a+b)=a^2-b^2[/itex]

    [itex](c-d)(c^2+cd+d^2)=c^3-d^3[/itex]

    Therefore, [itex]\ \ (\sqrt{x\,}-1)(\sqrt{x\,}+1)=x-1[/itex]

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

    These can be used to cancel x-1 in the numerator & denominator.
     
  5. Dec 20, 2012 #4
    Thanks, both responses were very helpful.
     
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