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Homework Help: Function under square root

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data

    lim x-> 2+ f(x)=sqrt(4-x^2)

    whats the value of the following function??

    2. Relevant equations

    3. The attempt at a solution

    i tried and got the answer as does not exist
    but some people got it as 0

    which is the correct answer
  2. jcsd
  3. Nov 4, 2013 #2


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    Can you explain the steps that lead to your result?
  4. Nov 4, 2013 #3


    Staff: Mentor

    f(2) = 0, but if x > 2, then the function is undefined. This means that the limit you showed does not exist.

    This limit does exist, however, and is equal to 0.
    $$\lim{x \to 2^-}\sqrt{4 - x^2}$$
  5. Nov 4, 2013 #4
    i assumed that x=2.1
    as we approach 2 from the right side

    the when i put it under the when we put it under the sqaure and subtract we get

    which is undefined
  6. Nov 4, 2013 #5
    but which is correct
    is my answer when x->2+ correct
  7. Nov 4, 2013 #6


    Staff: Mentor

    No. The limit as you wrote it doesn't exist.
  8. Nov 4, 2013 #7
    so im right
    it doesnt exist
  9. Nov 4, 2013 #8


    Staff: Mentor

    Yes. The limit doesn't exist.

    Your first post in this thread was confusing to me.
    I didn't understand that you were asking a question since you didn't end it with a question mark (?). I interpreted what you wrote as saying that 0 was the correct answer.
  10. Nov 4, 2013 #9
    sorry about that
    but anyway thanks for your ttime and help
    i owe you one
  11. Nov 4, 2013 #10


    Staff: Mentor

    You're welcome!
    That's OK. It's what we do here, and we enjoy doing it, as long as you make a reasonable effort when you post a question.
  12. Nov 4, 2013 #11

    Ray Vickson

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    In the real plane the limit does not exist because the function itself does not exist when x > 2. However, in the complex plane the function does exist, and the limit is 0. This is because for x > 2 we have
    [tex] 4 - x^2 = -(x^2 - 4) = (x^2 - 4) e^{\pm i \pi} \, \Longrightarrow
    \sqrt{4 - x^2} = \sqrt{x^2 - 4}\: e^{\pm i \pi/2} = \pm\, i \sqrt{x^2 - 4}.[/tex]
    The principal square root uses "+i", but the other is also a solution of z^2 = 4 - x^2. Anyway, no matter which square root you choose, it goes to zero along the imaginary axis in the complex plane, so the limit is zero.
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