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Limit of

  1. Sep 26, 2004 #1
    1) Compute the limit of f(x) = -(x-3)/square root of (x^2 - 9)
    as x approaches 3 from the right.

    2) Compute the limit of f(x) = - (x-3)/absolute value of (x-3)
    as x approaches 3 from the left.

    I got does not exist for both. Is that right?
     
    Last edited: Sep 26, 2004
  2. jcsd
  3. Sep 26, 2004 #2

    arildno

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    No that is incorrect.
    Let x>3
    Then
    [tex]\sqrt{x^{2}-9}=\sqrt{x-3}\sqrt{x+3}[/tex]
    Or:
    [tex]-\frac{x-3}{\sqrt{x^{2}-9}}=-\sqrt{\frac{x-3}{x+3}}[/tex]
    Hence, the right-hand side limit when x goes to 3 is 0.
    (The limit from the left-hand side doesn't make much sense for real f, since there are no x<3 which yields a real value for f(x)) (The domain of f(x) is restricted to x>3))
     
  4. Sep 26, 2004 #3

    arildno

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    Note:
    You've been asked to find the limit value when x approaches from one of the sides.
    Do not confuse this with the more usual question whether the function has a limit at a point (that is, a unique number so that irrespective which side you approach from, your evaluations tend to that number)
     
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