# Find the asymptotes

1. Jul 31, 2012

### frosty8688

1.Find the asymptotes of the following equation.

2. $\left.x/(x-1)^2\right.$

3. I know that the asymptote as x approaches ±∞ is ±∞. I am wondering when (x-1)$^{2}$ approaches ±∞ is also ±∞

2. Jul 31, 2012

### Staff: Mentor

To be an equation, there needs to be an = somewhere.
???
As x gets large or very negative, the denominator of x/(x - 1)2 is much larger than the numerator.

Now I think I know what you're trying to say. You seem to be using "asymptote" in place of "limit." The two words are not synonyms, and don't mean the same thing.

As x approaches ∞, both x and (x - 1)2 approach ∞. As x approaches -∞, x approaches -∞, but (x - 1)2 approaches +∞. Another way to say this is
$$\lim_{x \to -\infty} (x - 1)^2 = \infty$$

The thing is, the denominator gets large much more quickly than the numerator. One way to approach this problem is to factor x2 out of each term in the denominator, and factor x2 out of the numerator.

3. Jul 31, 2012

### frosty8688

So there would be no asymptotes.

4. Jul 31, 2012

### Staff: Mentor

Do you understand what an asymptote is? How would you define it in your own words?

5. Jul 31, 2012

### frosty8688

A vertical asymptote exists when x = a and a is determined by the values at which x is 0 and exists when f(x) = plus or minus infinity and a horizontal asymptote exists when x approaches infinity the number becomes larger and larger and if there is a denominator, the values when the x is in the denominator approach 0. So, there would be a horizontal asymptote at the x-axis. No vertical asymptotes, since the function approaches 0 as x goes to plus or minus infinity.

6. Jul 31, 2012

### eumyang

This is wrong. Take the denominator and set equal to 0 to find the vertical asymptote(s).

7. Jul 31, 2012

### frosty8688

Vertical asymptote at x = 1.

8. Aug 1, 2012

### HallsofIvy

Staff Emeritus
Yes, that is correct. Do you understand why eumyang said
?