# Thread: limit of a sequence

1. ## limit of a sequence

[img]http://img708.imageshack.us/img708/4042/analysis999.jpg[/img]
I stuck halfway through, any help would be appreciated.
-----------------------------------
Let ε>0 be given.
|x_n - (9/2)| = ......
simplified to:
39/[(4n^3) + 10]
Im stuck here, what do i do next?

2. Originally Posted by rlkmg
[img]http://img708.imageshack.us/img708/4042/analysis999.jpg[/img]
I stuck halfway through, any help would be appreciated.
-----------------------------------
Let ε>0 be given.
|x_n - (9/2)| = ......
simplified to:
39/[(4n^3) + 10]
Im stuck here, what do i do next?
Well you need to find$\displaystyle N$ in terms of $\displaystyle \epsilon$

Note that

$\displaystyle \displaystyle \bigg|\frac{39}{4n^3+10} \bigg|\le \frac{39}{4}\bigg|\frac{1}{n^3} \bigg|$

Now set this equal to epsilon to find big $\displaystyle N$

$\displaystyle \displastyle \frac{39}{4}\frac{1}{N^3} =\epsilon \iff N^3=\frac{39}{4\epsilon} \iff N=\sqrt[3]{\frac{39}{4\epsilon}}$

Now you can start the formal proof e.g

Let $\displaystyle \epsilon > 0$ and and let $\displaystyle N=\sqrt[3]{\frac{39}{4\epsilon}}$ then for $\displaystyle n > N$...

3. Originally Posted by TheEmptySet
$\displaystyle \displaystyle \bigg|\frac{39}{4n^3+10} \bigg|\le \frac{39}{4}\bigg|\frac{1}{n^3} \bigg|$
Thanks so much, this is the part that I dont understand, could you explain how you got from the left to the right of the above equation? I've seen your technique in many examples but i still dont understand how you did the above step. especially getting rid of the 10

4. Originally Posted by rlkmg
Thanks so much, this is the part that I dont understand, could you explain how you got from the left to the right of the above equation? I've seen your technique in many examples but i still dont understand how you did the above step. especially getting rid of the 10

$\displaystyle \displaystyle \frac{1}{12}=\frac{1}{2+10} \le \frac{1}{2}$