# Homework Help: Infinite Limits

1. May 8, 2010

### galwacco

1. The problem statement, all variables and given/known data
Hey people, my professor game us an assignment of dozens of limits to resolve. Most of them I managed to resolve, but two. I don't don't if it's something I can't resolve because I need more practice but it seems that I am really stuck with them. So if you can give me some direction with the first equation, I'd really appreciate.

2. Relevant equations
The equation is:

lim x$$\rightarrow$$$$\infty$$$$\frac{5x^6-x^5+4x^3-12}{2-x^2+8x^6}$$

3. The attempt at a solution

First, I tried to apply L'Hospital's theorem, but wasn't lucky with that, because it would yeld another equation just as complicated, so I asserted that it wasn't the way. Then I tried to divided the entire thing with the highest exponent X, which is x^6, but I got stuck, perhaps because I didn't really know how to make the division.

Anybody out there to help me?

2. May 8, 2010

### physicsman2

I don't see a trick to this problem, so do you know the properties of taking a limit at infinity?

You can observe the powers in the numerator and denominator and divide each term in the by the highest power present in the denominator. If the powers are the same in both num. and den., you should get a number. If the power is greater in the denominator, then your answer is usually zero. If the power is greater in the numerator, then you're going to have to do some synthetic division.

3. May 8, 2010

### Staff: Mentor

This is close to the right idea. Factor x^6 out of each term in the numerator, and each term in the denominator. It might be helpful to write the denominator with highest powers first. That will give you
$$\frac{x^6(5 + \text{other stuff})}{x^6(8 + \text{other stuff})}$$

Then take the limit. For any finite value of x, x^6/x^6 = 1, and this is true also in the limit.

4. May 13, 2010

### galwacco

Thanks a lot for your help.

I found out that if I divide it just like you said, by x^6, it boils the equation down to 5x^6/8x^6, which in turn will become 5/8! Which is the answer I was looking for.

Thanks again!

5. May 13, 2010

### Tedjn

That's by far the best way to do it. Just as an aside, l'Hopital's rule would work, but you would need to do it 6 times.

6. May 13, 2010

### galwacco

Exact Tedjn!
When I tried to use L'Hospital's rule, it didn't help much as I need to redo it many times. Yesterday I was taking a look at a specific equation that looked just like the one I was talking about, and I learned therein that any A/x tends to zero.

So, after dividing by x^6, everything I had left was the 5x^6/8x^6!
Pretty interesting! I am starting to like this calculus thing.

thanks again.