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Left & Right Limits Of A Function (Finding Values Of A Variable)

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data
    11rddt5.jpg


    2. Relevant equations
    Only ones I know that are relevant are the equations shown in the question (for the left and right limits).


    3. The attempt at a solution
    The difficulty that I am having right not is comprehending what is really being asked from the question. From what I understand, then question is asking the "m" values from the left and right limits at the top, and the value for "m" when those limits are the same. However, how would I go about finding the value of "m" if I have no x or y co-ordinates given?
     
  2. jcsd
  3. Sep 9, 2012 #2

    SammyS

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    What is [itex]\displaystyle \lim_{x\to\,1}(8x^3-2m)\ ?[/itex]
     
  4. Sep 9, 2012 #3
    That is not given (everything above that I posted is the only thing given). Or are you suggesting that I should calculate that to solve this problem. That expression just defines the limit from the left side. I am unsure on how to calculate that.
     
  5. Sep 9, 2012 #4

    SammyS

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    What do you mean, it's not given?

    I'm asking you to evaluate the limit, [itex]\displaystyle \lim_{x\to\,1}(8x^3-2m)\,,[/itex] if you can. If you can't, I don't how you can do this problem at all.
     
  6. Sep 9, 2012 #5

    I don't see how I can evaluate that limit... I could isolate "m" though by temporarily assigning "f(x)" as "y". Then I could isolate "m" in terms of "y" => "m(y)".
     
  7. Sep 10, 2012 #6

    Mark44

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    Sammy is not asking you to solve for m (i.e., isolate me) - just evaluate the limit, which by the way should be
    $$\lim_{x \to -1^{-}}(8x^3 - 2m) $$

    Your answer should be in terms of m.
     
  8. Sep 10, 2012 #7

    SammyS

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    [itex]\displaystyle \lim_{x\to\,1}(8x^3-2m) = 8(1)^3-2m=8-2m[/itex]

    That's all there is to it !
     
  9. Sep 10, 2012 #8
    I can't believe how easy this question actually was... I was making it too complex due to the fact that I wasn't comprehending this question very well. A sincere thank you to Mark and Sammy for their assistance. I really appreciate it.
     
    Last edited: Sep 10, 2012
  10. Sep 10, 2012 #9

    SammyS

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    So, what are you results?
     
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