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Limit of rational function to rational power

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the limit, WITHOUT using l'Hôpital's rule:

    [tex] \lim_{x \rightarrow -1} \frac{x^{1/3} + 1}{x^{1/5} + 1} [/tex]


    2. Relevant equations



    3. The attempt at a solution

    I tried to use the conjugate method which does not produce a useful outcome:

    [tex] \underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{1/3}}+1}{{{x}^{1/5}}+1}\left( \frac{{{x}^{1/3}}-1}{{{x}^{1/3}}-1} \right)=\underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{2/3}}-1}{{{x}^{8/15}}-{{x}^{1/5}}+{{x}^{1/3}}-1} [/tex]

    Thank you for your help!
     
  2. jcsd
  3. Oct 22, 2009 #2
    [tex]u^{15} = x, ~then ~u^3 = x^\frac{1}{5} ~and ~u^5 = x^\frac{1}{3}[/tex]

    Then factor and cancel.
     
  4. Oct 23, 2009 #3

    HallsofIvy

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    That's not the "conjugate method". [itex]x^a- 1[/itex] is not always "conjugate" to [itex]x^a+ 1[/itex]. A and B (both including roots) are "conjugate" if and only if AB does not have any roots. [itex]x^{1/2}- 1[/itex] is conjugate to [itex]x^{1/2}+ 1[/itex] because [itex](x^{1/2}- 1)(x^{1/2}+ 1)= (x^{1/2})^2- 1^2= x- 1[/itex] but [itex]x^{1/3}+ 1[/itex] is conjugate to [itex]x^{2/3}- x^{1/3}+ 1[/itex] because [itex](x^{1/3}+ 1)(x^{2/3}- x^{1/3}+ 1)= x+ 1[/itex]

    Instead use the facts that [itex]x+ 1= (x^{1/3}+ 1)(x^{2/3}- x^{1/3}+ 1)[/itex] and that [itex]x+ 1= (x^{1/5}+ 1)(x^{4/5}- x^{3/5}+ x^{2/5}- x^{1/5}+ 1)[/itex]. Those are the "conjugates".
     
  5. Oct 23, 2009 #4
    Thank you for your responses, Bohrok and HallsofIvy.

    @HallsofIvy: Thanks for your clarification on conjugates.

    How would you define a real conjugate then? Are two terms [tex] x [/tex] and [tex] y [/tex] conjugates of each other if and only [tex] x \times y [/tex] are of degree 1 and do not have any fractional exponents?
     
  6. Oct 23, 2009 #5

    HallsofIvy

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    Not necessarily of degree one but "does not have fractional exponents", yes.
     
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