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Limit as it approaches 0

  1. Mar 7, 2015 #1
    Why is the limit not just infinity?

    00kP8.jpg

    wouldn't it be (1-infinity)/(1+infinity)?
     
  2. jcsd
  3. Mar 7, 2015 #2

    mfb

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    Why should "-infinity/infinity" be infinity?

    What is the limit of
    $$\lim_{s \to 0} \frac{\frac{-1}{2}}{\frac{2}{s}}$$?
     
  4. Mar 7, 2015 #3
    -s/4, if s =0 then
    Would't the limit be zero?
     
  5. Mar 7, 2015 #4

    mfb

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    Sorry typo, this is the formula I meant:
    $$\lim_{s \to 0} \frac{\frac{-1}{s}}{\frac{2}{s}}$$

    But the more important part was the first question.
     
  6. Mar 7, 2015 #5
    For that formula it would be -1/2. I was thinking it would be infinity because you really can't put down a number with it unless maybe it is 1?

    EDIT: Nvm go it. It ends up being -35/40. Thnx
     
    Last edited: Mar 7, 2015
  7. Mar 8, 2015 #6

    mfb

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    That is not the point. It follows the "-infinity/infinity" type. It is easy to simplify it here to see the limit is not infinity, so your original idea cannot work - that was the purpose of the example.
    I don't understand that question.
    Okay.
     
  8. Mar 8, 2015 #7

    PeroK

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    Whenever you write "infinity", you should write "a large number".

    In this form (1 - a large number)/(1 + another large number) could be any negative number. It could be large and negative or it could be small and negative.
     
  9. Mar 8, 2015 #8

    HallsofIvy

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    A very important property of limits, not emphasized enough as I think it should be, is that "If f(x)= g(x) for all x except x= a, then [itex]\lim_{x\to a} f(x)= \lim_{x\to a} g(x)[/itex].

    With this particular problem, we can, as long as n is not 0, multiply both numerator and denominator by n. That gives you [tex]\frac{s- 1}{s+ \frac{s+ 4}{7(s+ 5)}}[/tex]. Now, take s= 0.
     
  10. Mar 8, 2015 #9

    mfb

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    You lost two prefactors, HallsofIvy.

    $$\frac{s- 1}{s+ \frac{10(2s+ 4)}{7(s+ 5)}}$$
     
  11. Mar 8, 2015 #10

    HallsofIvy

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    Thanks.
     
  12. Mar 9, 2015 #11
    The limit is of (s-1)/(s+(10*(2s+4)/7(s+5) that doing s=0 result in -7/15
     
  13. Mar 9, 2015 #12

    Mark44

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    No, check your algebra. The answer in post #1 (-7/8) is correct.
     
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