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The limit of finite approximations of area

  1. Jan 31, 2012 #1
    Screenshot2012-01-31at11045AM.png

    My textbook never mentioned what happens when you multiply something by infinity. I would think 4 * ∞ would be ∞. So to me that whole equation should simplify to 1 - ∞ which is ∞. I don't see how they get 2/3
     
  2. jcsd
  3. Jan 31, 2012 #2
    you're forgetting that we'd also have an '∞' downstairs too
    let's take this one step at a time

    [itex]1-\frac{2n^3+3n^2+n}{6n^2}[/itex]

    How can we simplify this thing first

    [itex]1-\frac{2n^3+3n^2+n}{6n^2}=1-\frac{2n^3}{6n^3} - \frac{3n^2}{6n^3} - \frac{n}{6n^3}=1-\frac{2}{6} - \frac{3}{6n} - \frac{1}{6n^2}[/itex]

    Now, what happens here as we take n away up to infinity?
    All of those [itex]\frac{1}{n}[/itex] terms go to zero, correct?

    So we are left with

    [itex]Limit_{n \rightarrow \infty} (1-\frac{2n^3+3n^2+n}{6n^2}) = Limit_{n \rightarrow \infty}(1-\frac{2}{6} - \frac{3}{6n} - \frac{1}{6n^2})[/itex]

    From there we can see the same result your books example gave you

    [itex]Limit_{n \rightarrow \infty}(1-\frac{2}{6} - \frac{3}{6n} - \frac{1}{6n^2})=1-\frac{2}{6}=\frac{6-2}{6}=\frac{4}{6}=\frac{2}{3}[/itex]
     
  4. Jan 31, 2012 #3

    Ray Vickson

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    Why do you think that [tex]\frac{2n^3 + 3n^2 + n}{6n^3} \rightarrow \infty[/tex] as [itex] n \rightarrow \infty ?[/itex] It doesn't, and I really cannot see why you would ever think it does. Note that [tex] \frac{2n^3 + 3n^2 + n}{6n^3} = \frac{2n^3}{6n^3} + \frac{3n^2}{6n^3} +
    \frac{n}{6n^3} = \frac{1}{3} + \frac{1}{2n^2} + \frac{1}{6n^2} [/tex] for any n > 0.

    RGV
     
  5. Jan 31, 2012 #4
    thanks i got it now
     
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