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Thread: Find the limit

  1. #1
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    Find the limit

    (x^(1/3)-1)/(x^(1/4)-1) find the limit as x approaches 1
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    Re: Find the limit

    (x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)=x-1 so you can multiply numerator and denominator by the second factor to simplify the numerator.

    You can do a similar trick with the denominator.
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    Re: Find the limit

    Quote Originally Posted by GodSalamence View Post
    (x^(1/3)-1)/(x^(1/4)-1) find the limit as x approaches 1
    I have never been a fan of telling a student to use L'Hopital's Rule. But in this case it seems justified, especially if one is algebraically challenged.

    $\displaystyle{{\lim _{x \to 1}}\dfrac{{{x^{(1/3)}} - 1}}{{{x^{(1/4)}} - 1}}\overbrace = ^H{\lim _{x \to 1}}\dfrac{{(1/3){x^{( - 2/3)}}}}{{(1/4){x^{( - 3/4)}}}}}$
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    Re: Find the limit

    Quote Originally Posted by Archie View Post
    (x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)=x-1 so you can multiply numerator and denominator by the second factor to simplify the numerator.

    You can do a similar trick with the denominator.
    I know you make it a practice to avoid using L'Hopital's Rule in doing these limit problems, but what you recommend
    here is relatively awkward, especially because of the fractional exponents. Let's still avoid L'Hopital's Rule.

    12 is the LCD of the fractions used for the exponents.


    Let  x = y^{12}.

    Then x^{1/3} = y^4.

    And x^{1/4} = y^3.


    As x approaches 1, y also approaches 1.


    \displaystyle\lim_{y\to 1} \dfrac{y^4 - 1}{y^3 - 1} =

    \displaystyle\lim_{y\to 1} \dfrac{(y - 1)(y^3 + y^2 + y + 1)}{(y - 1)(y^2 + y + 1)} =
    Last edited by greg1313; Sep 9th 2017 at 08:06 AM.
    Thanks from Archie
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    Re: Find the limit

    Quote Originally Posted by greg1313 View Post
    As x approaches 1, y also approaches 1.
    It's a minor point, but one that is usually missed and is occasionally important: the implication here is in the wrong direction. You are using the existence (and value) of the limit in y to infer the existence (and value) of the limit in x. Thus the relevant implication is that as y \to 1, x \to 1^+. I guess we ought to do a little more work to prove the left-sided limit.
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    Re: Find the limit

    Quote Originally Posted by Archie View Post
    Thus the relevant implication is that as y \to 1, x \to 1^+.
    This bit is wrong. Of course y^{12} can be smaller than 1. But the general point still stands.
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    Re: Find the limit

    If y = x^{1/12}, then as x approaches 1, so that will mean that y
    will approach 1, as they are equal.
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    Re: Find the limit

    As I said, it's a minor point in this case. But if the limit were zero, we'd have a problem because the equality x=y^{12} would hold only on one side of the limit point.
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    Re: Find the limit

    Quote Originally Posted by Archie View Post
    As I said, it's a minor point in this case. But if the limit were zero, we'd have a problem because the equality x=y^{12} would hold only on one side of the limit point.
    But the limit is not "as x approaches zero," so why bring that up? That doesn't have a bearing on what I'm doing.
    I already knew/know as x approaches 1 from the left and from the right, that x^{1/12} approaches 1.
    Last edited by greg1313; Sep 11th 2017 at 01:30 PM.
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