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Question on limits

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi everyone I just have this question in my exam, it's a limit question and I don't know how to solve it. Any help are much appreciated :D

    2. Relevant equations
    http://img42.imageshack.us/img42/8578/47934321.jpg [Broken]

    Find Lim (x -> -2) f(x) if it exist

    (sorry I don't know how to type the code)

    3. The attempt at a solution

    The answer I gave was the limit does not exist. I din't give an explanation for that.

    Thanks again for any help :D
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 24, 2009 #2

    LCKurtz

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    Look at the one sided limits as x --> -2 from the left and right.
     
  4. Nov 24, 2009 #3
    Ok I looked at the graph, and both the curve and line meet at x = -2 and y value is 9, do does that means that the limit of the function exist and it's at y = 9?
     
  5. Nov 24, 2009 #4

    LCKurtz

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    Yes. If the right and left limits at a point exist and are equal, then their common value is the limit of the function. On an exam you would likely be expected to give some justification for your conclusion that the right and left limits are both 9.
     
  6. Nov 25, 2009 #5
    Ok thanks for your help. I jhave one more question on the question itself. Because for the x^2 + 5, it was given that x is less than x is less than -2, what does that actually means? Thanks.
     
  7. Nov 25, 2009 #6

    HallsofIvy

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    As LCKurtz said initially, look at the "right" and "left" limits. To the left of x= -2, the function is just [itex]x^2+ 5[/itex]. That is a polynomial and so continuous for all x. In particular it is continuous at x= -2 and so its limit, as x approaches -2, is the value of the function there, [itex](-2)^3+ 5= 9[/itex].
    [tex]\lim_{x\to -2} X^2+ 5= 9[/itex].
    [tex]\lim_{x\to -2^-} f(x)= 9[/itex].

    To the right of x= -2, the function is 3- 3x. Again, that is a polynomial. It is also continuous at x= -2 and so
    [itex]\lim_{x\to -2} 3- 3x= 3- 3(-2)= 9[/itex]
    [itex]\lim_{x\to -2^+} f(x)= 9[/itex].

    Since the two one-sided limits exist and are equal, the limit itself exists and is that common value, 9.
     
  8. Nov 25, 2009 #7
    Ok thanks for the help, I will try to figure it out the whole thing again. Thanks again.
     
  9. Nov 27, 2009 #8
    I find it helpful to remember that a limit is not necessarily a tangible thing. As you saw in the function you were given there is a "hole" in the graph so -2 doesn't really exist, but the limit is in fact there. When you are looking for the limit, you are just looking for where the function would be if it actually existed at that point, whether it actually does exist there or not.

    Good luck!
     
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