1. Horizontal Asymptote Problem

I understand fairly well how to find horizontal asymptotes, but this function is really messing with me. I don't think it has any, but I want to be sure. If it does have some, I'm sure I'm missing some basic step. Anyways, here it is:

f(x) = 3x^3 - x +1 / x+3

What would be the horizontal asymptote(s) if they even exist for this function?

2. Is this the correct function?

$f(x) = \frac{3x^3-x+1}{x+3}$

To determine the end-behavior (i.e., what happens at the extreme "left" and extreme "right" of the graph), you can take the largest power term from each the numerator and denominator, and calculate the limit like this:

$\lim_{x\to+\infty} \frac{3x^3-x+1}{x+3} = \lim_{x\to+\infty} \frac{3x^3}{x} = \lim_{x\to+\infty} 3x^2 = +\infty$

You should also check the left side if it's not a simple rational function. In this case it's not necessary, but just to be thorough:

$\lim_{x\to-\infty} \frac{3x^3-x+1}{x+3} = \lim_{x\to-\infty} \frac{3x^3}{x} = \lim_{x\to-\infty} 3x^2 = +\infty$

Since the end-behavior goes to infinity on both sides, there aren't any horizontal asymptotes.

3. Originally Posted by drumist
Is this the correct function?

$f(x) = \frac{3x^3-x+1}{x+3}$

To determine the end-behavior (i.e., what happens at the extreme "left" and extreme "right" of the graph), you can take the largest power term from each the numerator and denominator, and calculate the limit like this:

$\lim_{x\to+\infty} \frac{3x^3-x+1}{x+3} = \lim_{x\to+\infty} \frac{3x^3}{x} = \lim_{x\to+\infty} 3x^2 = +\infty$

You should also check the left side if it's not a simple rational function. In this case it's not necessary, but just to be thorough:

$\lim_{x\to-\infty} \frac{3x^3-x+1}{x+3} = \lim_{x\to-\infty} \frac{3x^3}{x} = \lim_{x\to-\infty} 3x^2 = +\infty$

Since the end-behavior goes to infinity on both sides, there aren't any horizontal asymptotes.
Okay, I thought so, just wanted to make sure I wasn't missing something obvious.

Thanks for the help, I appreciate it.