# Limit Proof

1. Jun 2, 2008

### tvguide123

1. The problem statement, all variables and given/known data

Prove that the limit as x->inifinity [x^2 - 2x] / [x^3 - 5] = 0

2. Relevant equations

The general procedure that we have to use to come up with this proof is:
"For all epsilon>0, there exists some N>0, such that for all x, if x>N then this implies that
| [[x^2 - 2x] / [x^3 - 5] - 0] | < epsilon".

N and epsilon are just variables.

3. The attempt at a solution

I simplified the "| [[x^2 - 2x] / [x^3 - 5] - 0] |" down to
"| [[x^2 - 2x] / [x^3 - 5] |"

I'm allowed to manipulate this equation as long as it is preserved. Also I can make helper assumptions as to the value of "N" as long as I account for them in my proof.

I've spent hours staring at this question but I cant figure out how to proceed from here
Any help would be greatly appreciated!

2. Jun 2, 2008

### spideyunlimit

[x^2 - 2x] / [x^3 - 5]
divide by x^2 in the numerator and denominator:
[1 - (2/x)] / [x - (5/x^2)]
now use x tends to infinity in the above expression:
[1 - 0] / [ infinity - 0]
= 1 / infinity
= 0

3. Jun 2, 2008

### HallsofIvy

Staff Emeritus
First, have you thought about how you would have found limit itself- ignoring, for the moment the "first principes" stuff? Since x is going to infinity, what you have been taught to do is to divide both numerator and denominator by the highest power of x: in this case x3. That gives
$$\frac{\frac{1}{x}- \frac{2}{x^2}}{1- \frac{5}{x^3}}$$
The point of that is that "0" is much easier to work with than "infinity"! As x goes to infinity, every fraction with x in the denominator goes to 0:(0+ 0)/(1- 0)= 0.

Now, that is basically what you want to do in the proof itself. As long as x is not 0 (and "going to infinity" it won't be) you have
$$|\frac{x^2- 2x}{x^3- 5}|=| \frac{\frac{1}{x}- \frac{2}{x^2}}{1- \frac{5}{x^3}}|= |\frac{1}{x}\frac{1- \frac{2}{x}}{1- \frac{5}{x^3}}|$$

That first 1/x is what is crucial. As x goes to infinity, it will go to 0. As long as the remaining fraction is bounded making that less than $\epsilon$ will depend entirely on 1/x.

Choose a convenient value to make that fraction- say "1". How large must x be in order that
$$|\frac{1- \frac{2}{x}}{1- \frac{5}{x^3}}< 1|$$?
Once you have determined that, take x> N1> "that number" and you have reduced the problem to
$$|\frac{x^2- 2x}{x^3- 5}|\le \frac{1}{x}< \epsilon$$
and you can easily find an N2 that will make that true. Take N to be the larger of N1 and N2 so that they are both true.

4. Jun 2, 2008

### spideyunlimit

Such a complex explanation for something so simple? :D
just remember that any number with denominator tending to infinity tends to 0.

5. Jun 3, 2008

### Gib Z

Do you know what epsilon-delta proofs are? When something asks to prove a limit, it means using an epsilon-delta proof. If it asks to show that a limit tends to some value, then you may do it as you did.

6. Jun 3, 2008

### HallsofIvy

Staff Emeritus
That was what I said wasn't it? Except that the OP asked for a PROOF so I based a proof on that idea.

7. Jun 8, 2008

### tvguide123

Sorry spidey we had to be more rigorous in our proof.

Thanks a bunch HallsofIvy that got me thinking on the right track although it wasnt exactly in the correct method that we have to do it in.

Yes gibz we had to structure it as an epsilon-delta proof but i was able to use hallsofivys solution to create my own

Thanks guys Problem Solved :)

8. Sep 14, 2008

### qwertyuiopasd

Spidey you are wrong. e[x][/SUP]/x goes to infinity. The denominator goes to infinity also but the limit is not 0!!!