1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: The limit of finite approximations of area

  1. Jan 31, 2012 #1

    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


    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

    User Avatar
    Science Advisor
    Homework Helper

    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.

  5. Jan 31, 2012 #4
    thanks i got it now
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook