Homework Help: The limit of finite approximations of area

1. Jan 31, 2012

bobsmith76

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. Jan 31, 2012

genericusrnme

you're forgetting that we'd also have an '∞' downstairs too
let's take this one step at a time

$1-\frac{2n^3+3n^2+n}{6n^2}$

How can we simplify this thing first

$1-\frac{2n^3+3n^2+n}{6n^2}=1-\frac{2n^3}{6n^3} - \frac{3n^2}{6n^3} - \frac{n}{6n^3}=1-\frac{2}{6} - \frac{3}{6n} - \frac{1}{6n^2}$

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

So we are left with

$Limit_{n \rightarrow \infty} (1-\frac{2n^3+3n^2+n}{6n^2}) = Limit_{n \rightarrow \infty}(1-\frac{2}{6} - \frac{3}{6n} - \frac{1}{6n^2})$

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

$Limit_{n \rightarrow \infty}(1-\frac{2}{6} - \frac{3}{6n} - \frac{1}{6n^2})=1-\frac{2}{6}=\frac{6-2}{6}=\frac{4}{6}=\frac{2}{3}$

3. Jan 31, 2012

Ray Vickson

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

RGV

4. Jan 31, 2012

bobsmith76

thanks i got it now

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook