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Question is a VA exists

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    If f(-6) = 3, is it possible for there to be a vertical asymptote at x=-6?
    The limit as x approaches -6 is positive infinity, but there f(-6) is defined by the value of -3. Or, since VAs refer to the line, the fact that f(-6) doesn't matter?
     
  2. jcsd
  3. Sep 17, 2009 #2

    Mark44

    Staff: Mentor

    How f is defined at x = -6 doesn't matter.

    Here is an example function that has the behavior you describe.

    f(x) = 1/(x + 6), if x != -6
    f(x) = 3, if x = -6

    This function has a vert. asymptote at x = -6, even though f(-6) is defined.
     
  4. Sep 17, 2009 #3
    Thanks!
    Another question:
    For the piecewise,
    f(x){-x+1 , 0<=x<1
    {1, 1<=x<2
    What is the limit as x approaches 1?

    Do I do?
    -x+1 = 1
    -x=0
    x=0?
    So 0+1 = 1
    and 1
    So the limit is 1? Or does it not exist?
     
  5. Sep 17, 2009 #4

    Mark44

    Staff: Mentor

    For a limit to exist, both one-sided limits must exist and must be equal.
    So you need to find these limits:
    [tex]\lim_{x \rightarrow 1^+}f(x)[/tex]
    [tex]\lim_{x \rightarrow 1^-}f(x)[/tex]
    If both exist and are equal, then
    [tex]\lim_{x \rightarrow 1}f(x)[/tex]
    exists and is equal to the common value.
     
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