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Limit x->1 of [sqrt(x-1)]

  1. Sep 15, 2011 #1
    limit x-->1 of [sqrt(x-1)]

    i know that for limit to be defined the left and right handed limits should be equal.
    in the above problem, the left handed limit is undefined because it is the square rot of a negative no. but i checked this question on alphawolfram.com and also on MATHEMATICA . both the programs were giving limit as 0
    so i supposed i was missing something but when i submitted my answer as 'undefined' it was marked as correct and i was given full credit
    so please everyone clarify this problem
    and also consider this one: lim x-->1 of sqrt[1-x]
    in this case the right hand liit is not defined but again mathematica gave the same answer '0'
     
  2. jcsd
  3. Sep 15, 2011 #2

    Hootenanny

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    Re: limit x-->1 of [sqrt(x-1)]

    You are correct in saying that for [itex]f:\mathbb{R}^+\mapsto\mathbb{R}^+[/itex], [itex]f:x\mapsto\sqrt{x}[/itex], the left-hand limit does not exist. However, Mathematica and Wolfram Alpha (and Maple, I would imagine as well as most CAS) will quite happily work in the complex plane, where the limit does exist and is indeed zero.
     
    Last edited: Sep 15, 2011
  4. Sep 15, 2011 #3
    Re: limit x-->1 of [sqrt(x-1)]

    yes you are right there.
    on the graph one of the lines was marked as real and other as imaginary.
    so it means as far as i have learned, my answer was correct. the limit for both the functions is un defined
     
  5. Sep 15, 2011 #4

    Hootenanny

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    Re: limit x-->1 of [sqrt(x-1)]

    Both functions, there is only one function: the square root. The two lines on the graph indicate the values of the function (both the real and imaginary parts).
     
  6. Sep 15, 2011 #5
    Re: limit x-->1 of [sqrt(x-1)]

    ok...but i have not been taught any complex valued functions yet.
    so as far as my assingnment is concernerd i should submit this as undefined.
    thanks a lot
     
  7. Sep 15, 2011 #6

    Hootenanny

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    Re: limit x-->1 of [sqrt(x-1)]

    It would seem so, yes.
    No problem.
     
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