The limit of finite approximations of area

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  • #1
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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
 

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  • #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]
 
  • #3
Ray Vickson
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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
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
 
  • #4
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thanks i got it now
 

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