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Conjugate the limit

  1. Feb 21, 2009 #1
    $\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$

    1. The problem statement, all variables and given/known data
    Calculate the limit of [tex]\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}[/tex].

    2. Relevant equations
    As above.

    3. The attempt at a solution
    Have tried to multiplicate with the conjugate.
     
  2. jcsd
  3. Feb 21, 2009 #2

    cristo

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    Re: $\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$

    Ok, what did you get? Note that you must show your work in order to get help here.
     
  4. Feb 21, 2009 #3
    Re: $\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$

    [tex]\lim_{x\to 1} \frac{\sqrt{x}-1}{x-1} = \lim_{x\to 1} \frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1)}{\left(x-1\right)\left(\sqrt{x}+1\right)} = \lim_{x\to 1} \frac{x-1}{x\sqrt{x}+x-\sqrt{x}-1} = \lim_{x\to1}\frac{x-1}{\sqrt{x}\left(x-1\right)+x-1} = \lim_{x\to1}\frac{x-1}{(x-1)(\sqrt{x}+1)} = \frac{1}{2}[/tex]

    So in essence, disregard me, for I am retarded. :P
     
  5. Feb 21, 2009 #4

    HallsofIvy

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    Re: $\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$

    For a retarded person, remarkably good at limits!
     
  6. Feb 21, 2009 #5
    Re: $\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$

    How about LHopital's rule?
     
  7. Feb 21, 2009 #6

    HallsofIvy

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    Re: $\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$

    Why? That's like using a sledgehammer to crack a walnut. Walker242's solutions is excellent- especially because it is his solution!
     
  8. Feb 21, 2009 #7
    Re: $\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$

    Why?
    His solution is very good.
    So I have no overriding reason; but LHopital is more generic.
    But I am into generic, versus tricky.
     
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