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Computing these limits?

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Just checking if these are right?

    f(b) = 2-2√b
    compute limit in are as:
    b-> 0+
    b->1-
    Explain in a brief sentence why it does not make sense to compute a limit as b->0-


    2. Relevant equations
    given above




    3. The attempt at a solution
    lim b->0+ (2-2√b) = 2
    lim b->1- (2-2√b) = 0
    Lim b->0- (2-2√b) = ? The question says it doesn't make sense to do so. Is it not 2?
     
  2. jcsd
  3. Jan 28, 2013 #2

    Dick

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    Fine for the first two. For the last if b->0- then b is negative. If you are working in the real numbers then taking the square root of a negative number should make you feel odd at least. Why? This is another example of where blind plugging without thinking is not the best idea.
     
  4. Jan 28, 2013 #3
    Thank you! Is it because square root of a negative is "i"
     
  5. Jan 28, 2013 #4

    SammyS

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    The square root of -1 is i .

    [itex]\displaystyle \sqrt{-|a|}=i\cdot\sqrt{|a|}[/itex]
     
  6. Jan 29, 2013 #5

    Dick

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    The square root of -1 is also -i, but main point is that if you are working with real numbers and limits the square root of a negative number is simply undefined. It's not real. I think that's what the are expecting you to say. Can you tell me why there is no real number whose square is negative?
     
  7. Jan 29, 2013 #6
    I should of been more clear in terms of stating it being undefined. Sorry
     
  8. Jan 29, 2013 #7

    Dick

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    I wasn't looking for an apology and there is none needed. I was just looking you to say how you would explain that there was no limit b->0-.
     
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