# Using limits to find asymptotes.

1. Jan 3, 2014

### physics604

1. $$f(x)=x-\frac{1}{6}x^2-\frac{2}{3} lnx$$

2. Relevant equations

limits

3. The attempt at a solution

I know there is a vertical asymptote at x=0 because all values of x have to be greater than x.

The answer says that there is no horizontal asymptote, but I don't know how it reaches that conclusion.

Using limits, I can rewrite the equation as $$limx→∞ (\frac{x}{x^2}-\frac{1}{6}\frac{x^2}{x^2}-\frac{2}{3}\frac{lnx}{x^2})$$ which turns into $$limx→∞ (\frac{1}{x}-\frac{1}{6}-\frac{2}{3}\frac{lnx}{x^2})=-\frac{1}{6}$$ What's wrong with this logic?

2. Jan 3, 2014

### SteamKing

Staff Emeritus
IDK what you mean by the statement, "all values of x have to be greater than x." There seems to be some kind of logical inconsistency there.

Also, your limit doesn't evaluate to -1/6 as x goes to infinity. I believe the log term in your modified expression tends to infinity/infinity as x goes to infinity, which would indicate that you need to rush your function to the L'Hopital stat!

3. Jan 4, 2014

### physics604

Sorry, I meant all values of x have to be greater than 0.

I'm taking Calculus AB in school and we don't cover L'Hopital's Rule in that course. My modified function was the alternate method that my teacher taught us. He said that we should divide everything by the largest degree of x so that we would end up with all variables in the denominator (like in my modified function). So, when we input ∞ in, most of the fractions with x on the bottom would go to zero. Whatever we have left is our horizontal asymptote.

How would I find the horizontal asymptote using that method? I get -1/6 but there shouldn't be a HA.

4. Jan 4, 2014

### haruspex

You can extract it as a factor, but not throw it away. That is, your method should produce -x2/6. This represents the 'leading term' for large x. Since that does not have a horizontal asymptote, neither does the original function.

5. Jan 4, 2014

### Staff: Mentor

There is not a horizontal asymptote. The x2 term is the dominant term in your function. I'm not sure what the purpose of dividing by x2 is, other than possibly convincing you that it is the dominant term.

Why would you think there is a horizontal asymptote? In order of importance, from greatest to least, the terms are x2, x, and ln(x). For large x, your function looks like y = -(1/6)x2. The contributions of the other two terms are of minimal importance for large |x|.

6. Jan 4, 2014

### Mentallic

You're supposed to be taking the limit of f(x) as $x\to\infty$ but you've started working with f(x)/x^2. This is an entirely different function.

$$f(x)=x-\frac{1}{6}x^2-\frac{2}{3}\ln{x}$$
$$=\frac{x^2}{x^2}\left(x-\frac{1}{6}x^2-\frac{2}{3}\ln{x}\right)$$
$$= \frac{x^3-\frac{1}{6}x^4-\frac{2}{3}x^2\cdot\ln{x}}{x^2}$$

Notice the difference with what you've done? You simply multiplied the denominator by x2, hence changing the function, but if you multiply both the numerator and denominator by x2 then nothing has been changed (other than possibly having x not equal to 0).

But this isn't what you want to do for this problem. What you should be doing is what SteamKing suggested, but since your class isn't up to that, I'm sure you're allowed to assume that ln(x) grows slower than x. That is,

$$\lim_{x\to\infty}\frac{\ln{x}}{x}=0$$

Haha