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Simple limits question

  1. Sep 5, 2009 #1
    hey hi.

    i'm just starting differentials and now we are seeing the limits.
    i have a question.

    there is one example that goes like this :

    lim 4 / (x²+2x+1)
    x--> -1

    well with simple calculations using -1(-) and -1(+), i know that the answer is infinity.
    plus, if i trace the graph, there's an asymptote at -1.
    but i'm not supposed to use a calculator.
    well at the exam we can't have one.

    so i do it by hand.
    when i plug, let's say, -1.000001 and -0.999999, it gives me 4/0 for both.
    can i say that it's infinity, or would it have to be 4/0(+)?

    i feel that it's really easy, yet i'm stuck :(

    also, if i get something like (1(+))², does it stays 1(+) or does it becomes like 1.0000000002?

    thanks!
     
  2. jcsd
  3. Sep 5, 2009 #2

    tiny-tim

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    Hi gamma-ray-burst! :smile:
    hmm … rather long-winded :redface:

    rewrite it as a limε -> 0, where x = -1 + ε. :wink:
     
  4. Sep 5, 2009 #3

    CompuChip

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    In this case, you can "plug in" x = -1. Then you find that you get something of the form 4 / 0. Therefore, the limit does not exist; we sometimes write
    [tex]\lim_{x \to -1} \frac{4}{x^2 + 2x + 1} = \infty[/tex]
    to indicate this ([itex]\infty[/itex] being the symbol for infinity -- note that this is like a convention to say that the limit does not exist, it does not mean that the limit is "equal to infinity" or something like that).

    In principle, this method works: when you get something of the form number/0 or infinity/number (where in both cases, number is not equal to 0), the limit does not exist. Again, I should point out that "of the form" is a necessary addition here: technically you are not allowed to write "4/0", for example, that is an undefined expression. Only when you get something like 0/0 or infinity/infinity you really need to be careful and apply another method.
     
  5. Sep 5, 2009 #4

    HallsofIvy

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    Having noted that, at x= -1, the numerator is 4 and the denominator is 0, your answer should be that the limit does not exist.

    Strictly speaking,
    [tex]\lim_{x\to -1} \frac{4}{x^2+ 2x+ 1}= \infty[/tex]
    is not correct because
    [tex]\lim_{x\to -1^+} \frac{4}{x^2+ 2x+ 1}= \infty[/tex]
    while
    [tex]\lim_{x\to -1^-} \frac{4}{x^2+ 2x+ 1}= -\infty[/tex]

    Best answer: "The limit does not exist."
     
  6. Sep 5, 2009 #5
    Oh wow you guys are amazing.
    Thanks a lot for all the answers!
    Yet again, I was complicating things for nothing.
    I know that next week we will see indeterminations of forms like "0/0" and "inf/inf" and things like that and I was afraid that "#/0" was some special case where I had to do other things.

    So thanks again!
     
  7. Sep 5, 2009 #6
    This is untrue.

    [tex]{x^2+ 2x+ 1}[/tex] = [tex](x+1)^2. [/tex]

    Whatever side you approach, the square will make it positive, thus positive infinity.
     
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