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Solving an indeterminate 0/0 limit without L'Hospital

  1. Mar 17, 2012 #1
    Hi! I'd like to know how to solve the following limit without having to use L'Hospital's rule:

    [tex]\displaystyle\lim_{x \to{1}}{x^{\frac{1}{x-1}}}[/tex]

    I've already solved it using the derivative rules but I want to know how to solve it without them. Thanks for your help!

    EDIT: I was mistaken, it's not a 0/0 indetermiante limit, it's 1^infinity. Edit the title, thanks!
     
  2. jcsd
  3. Mar 17, 2012 #2
    Does the limit from the left equal the limit from the right? Also, l'Hospital's rule doesn't apply to this kind of limit. What kinds does[\I] it apply to?
     
  4. Mar 17, 2012 #3
    There's one point when you're solving this limit that you get an indeterminate 0/0 limit:

    [tex]e\displaystyle\lim_{x \to{1}}{\frac{log(x)}{x-1}}[/tex]

    When you get there you may apply L'Hospital's rule but I want to know how to continue without using it.
     
  5. Mar 17, 2012 #4

    HallsofIvy

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    Well, scurty, you can use L'Hopital, indirectly: If [itex]y= x^{1/(x-1)}[/itex] then [itex]ln(y)= (ln(x))/(x- 1)[/itex]which is of the form 0/0. Unfortunately, Hernaner28, I don't see any simple way to do this other than L'Hopital's rule.
     
  6. Mar 17, 2012 #5
    I'm confused, I thought the limit doesn't exist? The limit from the left doesn't equal the limit from the right.

    Edit: I'm on a mobile phone and I completely missed the x being raised to the power, I just saw the power. The x blended in with the lim. Okay, I understand what is being asked now.
     
  7. Mar 17, 2012 #6
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