# If |x| is large, what is f(x) approximately?

1. Nov 1, 2012

### dumbQuestion

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

If |x| is large, then f(x)=(x5-x4+x3+x)/(x3-1) is approximately what?

2. Relevant equations

Just use long division

3. The attempt at a solution

Well, I just started out dividing the polys, and I ended up with f(x)=x2-x+1 + (x2+1)/(x3-1)

I thought, well, if x is very large, then the fraction at the end there will begin disapearing and tend towards 0. The solution in the book agreed, but I'm confused about something. The solution in the book says "as the limit as x --> infinity, (x2+1)/(x3-1) = 0, so f(x) ≈x2-x+1 But this doesn't make sense because as x tends to infinity, x2-x+1 will blow up towards infinity. I guess I am just not wrapping my mind around how f(x) is approximately x2-x+1 for |x| "very large". Is it just because the fraction goes to 0 faster than x2-x+1 goes to infinity?

2. Nov 1, 2012

### bengunn

Essentially f(x) is infinity +0 to begin, because |x| is large. Work back from that assumption and it seems clear since the limit is infinity

3. Nov 1, 2012

### Simon Bridge

Well the book was being a bit sloppy - as x gets very large f(x) approaches infinity as the function y=x2-x+1.

It means:
If you plotted the whole f(x), and the quadratic y as a dotted line, you'll find that f(x) gets closer and closer to following the dotted line. That is how f(x) approaches infinity.

4. Nov 1, 2012

### LCKurtz

The graphs of f(x) and x2-x+1 will become very close together as x gets large, because the difference between them goes to zero. It doesn't matter that they are both going to infinity. They will still be close together as they do.

5. Nov 1, 2012

### bengunn

asymptote

Last edited: Nov 1, 2012