# Proof of a limit of a cubic like sequence.

1. Aug 12, 2008

### torquerotates

Now this is an example in my book that I don't really understand. It says prove that the limit of (4n^(3)+3n)/(n^(3)-6))=4 as n goes to infinity.

Basically the definition of a limit of a sequence as n->infinity is as follows.
Lim{a}=L

For any epsilion>0 there is a N>0 such that n>N=> |{a}-L|<epsilion

So here's the book's solution. I just don't know why they found upper bounds.

|(4n^(3)+3n)/(n^(3)-6) - 4|=|(3n+4)/(n^(3)-6)|<epslion

but if n>or=2, (3n+4)/(n^(3)-6)<epslion.

(3n+24)<or =30n

& (n^(3)-6)>or =(1/2)n^(3)

So, (3n+4)/(n^(3)-6)<or= (30n)/( (1/2)n^(3))<epsion

=> 60/n^(2)<epslion
=> (60/epslion)^(-1/2)<n

so this implies that we make N=max{2, (60/epslion)^(-1/2)}

Now here's my much more simplistic solution.

I never found a reason why they used an upper bound. Wouldn't a lower bound work just as well?

(3n+4)/(n^(3)-6)<epslion. { still given that n>or=2}

n/(n^(2)) <(3n+4)/(n^(3)-6)<epslion. Now I chose the denominator,n^2, arbitaritly just to make just sure that the denominator becomes smaller.

So solving, 1/epslion<n. This implies that we make N=max{2, 1/epsilion}.

See how simple this is. I didn't even use equal to or greater than/ less than. I just used a straightforward inequality. But is it correct? I have no clue why the book did it with upper bounds.

2. Aug 12, 2008

### Alex6200

(4n^(3)+3n)/(n^(3)-6))=4
(4 * oo^3 + 3*oo)/(oo^3-6)) = 4
(4 * oo^3)/(oo^3)
4 = 4

I don't see what would be wrong with just showing that kn^3/n^3 for all n is k, and that n^3 + n = n^3 as n approaches infinity.

I could be wrong though.

3. Aug 13, 2008

### HallsofIvy

Staff Emeritus
I'm not sure you can justify cancelling infinities like that. More rigorous is to divide by both numerator and denominator by the highest power of n:
$$\frac{4n^3+ 3n}{n^3- 6}= \frac{4+ \frac{3}{n^2}}{1- \frac{6}{n^3}}$$
Since both 3/n2 and 6/n3[/sub] go to 0 as n goes to infinity, the limit is 4/1= 4.

4. Aug 13, 2008

### torquerotates

But I'm trying to understand it from a formal approach using epslion and a bound as they do in real analysis.

5. Aug 13, 2008

### roam

For proving this limit I would use the same method as Hall, I would devide each term in the numerator and denominator by the highest power of x that occurs in the denominator, namely, n3. Just like he did.

You must remember that $$lim_{x \rightarrow\pm \infty} \frac{1}{x} = 0$$ etc, and voila, you have limit = 4

Why can't you try L'Hopital's rule, btw?

6. Aug 13, 2008

### torquerotates

I'm not allowed to do that because I'm self studying real analysis. I need to prove things from the most fundamental steps. I have to use the definition of a limit. Which is, for a sequence,

Lim{a}=L as n-> infinity means,

For any epsilion>0 there is a N>0 such that n>N=> |{a}-L|<epsilion