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Thread: Finding a limit without using L'Hopital's rule or Maclaurin Series

  1. #1
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    Finding a limit without using L'Hopital's rule or Maclaurin Series

    I know what the answer is (according to all resources I checked, it should be a). However I have no clue how to get to it without using L'Hopital.
    $$\lim_{x\to\1} \frac{x^a - 1}{x-1}$$ , $a\in$\mathbb{R}$ , a \geq 0

    Thanks in advance.

    Edit: should've probably said that a\in$\mathbb{R}$ , and is not just integers.
    Last edited by StereoBucket; Jan 6th 2017 at 04:41 PM.
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  2. #2
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    Re: Finding a limit without using L'Hopital's rule or Maclaurin Series

    Thanks, but I'm pretty sure no one would accept this as an answer as it assumes that a is an integer, when it may not be.
    Sorry for not stating that information. I edited the post to reflect on that.
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    Re: Finding a limit without using L'Hopital's rule or Maclaurin Series

    \frac{x^a-1}{x-1}=\frac{e^{a\log x}-1}{e^{\log x}-1}=\frac{e^{ah}-1}{h}\cdot\frac{h}{e^h-1}
    where h=\log x \to 0

    When we now take the limit, the first term is the derivative of e^{at} evaluated at t=0, while the second is the reciprocal of the derivative of e^t evaluated at t=0.
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  4. #4
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    Re: Finding a limit without using L'Hopital's rule or Maclaurin Series

    Quote Originally Posted by Archie View Post
    \frac{x^a-1}{x-1}=\frac{e^{a\log x}-1}{e^{\log x}-1}=\frac{e^{ah}-1}{h}\cdot\frac{h}{e^h-1}
    where h=\log x \to 0

    When we now take the limit, the first term is the derivative of e^{at} evaluated at t=0, while the second is the reciprocal of the derivative of e^t evaluated at t=0.
    How is this not equivalent to using one of the methods we were asked to avoid?
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  5. #5
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    Re: Finding a limit without using L'Hopital's rule or Maclaurin Series

    I haven't taken a derivative, I've recognised a difference quotient.
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