Very quick math question

This is a discussion on Very quick math question within the A Brief History of Cprogramming.com forums, part of the Community Boards category; <edit> Bah, after re-reading my post I guess it's not a very quick question. </edit> So I just took a ...

  1. #1
    Software Developer jverkoey's Avatar
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    Very quick math question

    <edit>
    Bah, after re-reading my post I guess it's not a very quick question. </edit>

    So I just took a midterm, and one of the questions was as follows:

    Code:
    lim      (1-x^2)
    x->+inf  -------
              (3+x)
    I found two ways to go about doing this.

    1) divide all by x:
    Code:
    lim      (1-x^2)/x
    x->+inf  -------
             (3+x)/x
    
    lim      (1/x-x)
    x->+inf  -------
             (3/x+1)
    When you plug in +inf, you get:
    Code:
    -inf
    ----
      1
    Evaluating to -inf.

    I checked this over at the end of the test though, and decided I didn't like this solution, as by the definition of a limit: the left side limit must equal the right side limit...and how do you evaluate the left and right side of infinity?

    So, I erased that answer and proceeded to write answer #2:

    Code:
    lim      (1-x^2)
    x->+inf  -------
              (3+x)
    
    As   lim    lim
       x->+inf  1/x->0.
    
    Therefore,
    
    lim    (1-(1/x)^2)
    1/x->0 ----------
            (3+(1/x))
    So, for lim 1/x->0+, we have:
    Code:
    -inf
    ----
    +inf
    Which is -inf.

    Also, for lim 1/x->0-, we have:
    Code:
    -inf
    ----
    -inf
    Which is +inf.

    Code:
    lim1/x->0- DNE lim1/x->0+,
    Therefore lim1/x->0 DNE.
    So, yah, a bunch of my friends all said -inf is right, but I really don't like the idea of putting that down as saying it is -inf implies the limit exists and approaches -inf, which isn't necessarily true, as inf isn't defined as a number....Aurgh.
    Last edited by jverkoey; 10-24-2005 at 08:14 PM.

  2. #2
    & the hat of GPL slaying Thantos's Avatar
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    Code:
    lim      (1-x^2)
    x->+∞  -------
              (3+x)
    Since you are talking about x as it gets really large (becomes unbounded) you can ignore the parts that don't contain x so you basically get:
    Code:
    lim      -x^2
    x->+∞  -------
              x
    which can then reduced to
    Code:
    lim      -x
    x->+∞
    then you plug in ∞ and you get -∞. At this point all you are saying is that as x becomes unbounded in the positive direction the results diverge in the negative direction. Since it doesn't converge to a number the limit does not exist.

  3. #3
    Software Developer jverkoey's Avatar
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    Thanks Thantos.

    Now I must search for that inf ascii value...

  4. #4
    & the hat of GPL slaying Thantos's Avatar
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    start -> run -> charmap -> advanced view -> search for "infi" -> search -> select -> copy -> alt+tab -> paste

  5. #5
    Cat Lover
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    the infinity sign isn't ascii, it'd be unicode, like delta and all that.

  6. #6
    Crazy Fool Perspective's Avatar
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    Thantor is correct. But also,

    In your reasoning above jver, you said a couple times that inf/inf = something, somthing. This isn't true, inf/inf is undefined. You need to apply L'hopitals's rule to evauluate it.

  7. #7
    Software Developer jverkoey's Avatar
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    Quote Originally Posted by Perspective
    Thantor is correct. But also,

    In your reasoning above jver, you said a couple times that inf/inf = something, somthing. This isn't true, inf/inf is undefined. You need to apply L'hopitals's rule to evauluate it.
    Ahh, yes, stupid mistake on my part. So I guess the first method is correct and just using the fact that it approaches infinity implies that it doesn't exist.

  8. #8
    aoeuhtns
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    Quote Originally Posted by jverkoey
    I checked this over at the end of the test though, and decided I didn't like this solution, as by the definition of a limit: the left side limit must equal the right side limit...and how do you evaluate the left and right side of infinity?
    You don't. lim x->infinity is different notation and has a mildly different meaning than lim x->c, where c is some real number.

  9. #9
    Bond sunnypalsingh's Avatar
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    Quote Originally Posted by jverkoey
    Ahh, yes, stupid mistake on my part. So I guess the first method is correct and just using the fact that it approaches infinity implies that it doesn't exist.
    I wouldn't say infinity doesn't exist...i would say it's unreachable.....
    Actually infinity is a comparitive term.....

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