# Limit

Printable View

• Nov 10th 2009, 03:46 PM
Em Yeu Anh
Limit
$\lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n} \displaystyle{[(}1+\frac{3i}{n}\displaystyle{)}^3-2(1+\frac{3i}{n}\displaystyle{)]}$
• Nov 10th 2009, 04:31 PM
adkinsjr
Quote:

Originally Posted by Em Yeu Anh
$\lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n} \displaystyle{[(}1+\frac{3i}{n}\displaystyle{)}^3-2(1+\frac{3i}{n}\displaystyle{)]}$

You should show some effort. We aren't suppose to just solve problems for you.
• Nov 10th 2009, 04:32 PM
redsoxfan325
Quote:

Originally Posted by Em Yeu Anh
$\lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n} \displaystyle{[(}1+\frac{3i}{n}\displaystyle{)}^3-2(1+\frac{3i}{n}\displaystyle{)]}$

I would assume you haven't done the fundamental theorem of calculus yet, because this Riemann sum appears to be $\int_1^4x^3-2x\,dx=\frac{195}{4}$

Pull the $\frac{3}{n}$ out and expand the summand. You get

$\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^n\left[(3i/n)^3+3(3i/n)^2+(3i/n)-1\right]$ $=\lim_{n\to\infty}\frac{3}{n}\left[\frac{27}{n^3}\sum_{i=1}^n i^3+\frac{27}{n^2}\sum_{i=1}^n i^2+\frac{3}{n}\sum_{i=1}^n i-\sum_{i=1}^n 1\right]$

Evaluate that using summation formulas and then take the limit.
• Nov 10th 2009, 05:22 PM
Em Yeu Anh
I have studied the fundamental thereom although it was quite some time ago, would you mind explaining how you obtained 4 on the upper limit of integration for $\int_1^4x^3-2xdx?$

Sorry, think I may have just found it:
$\frac{3}{n} = \frac{b-1}{n}$ is that the formula?
• Nov 10th 2009, 06:01 PM
redsoxfan325
Yup. The reason I thought you hadn't done the FTC yet was that I assumed you wanted to evaluate this limit in order to evaluate the integral. I guess not.
• Nov 10th 2009, 06:09 PM
Em Yeu Anh
Alright I got 195/4! :D Thank you very much redsoxfan you have been most helpful
• Nov 10th 2009, 06:15 PM
redsoxfan325
You're welcome.