## Find the asymptotes

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 ±∞

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 Quote by frosty8688 1.Find the asymptotes of the following equation. 2. $\left.x/(x-1)^2\right.$
To be an equation, there needs to be an = somewhere.
 Quote by frosty8688 3. I know that the asymptote as x approaches ±∞ is ±∞.
???
As x gets large or very negative, the denominator of x/(x - 1)2 is much larger than the numerator.

 Quote by frosty8688 I am wondering when (x-1)$^{2}$ approaches ±∞ is also ±∞
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.
 So there would be no asymptotes.

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## Find the asymptotes

 Quote by frosty8688 So there would be no asymptotes.
Do you understand what an asymptote is? How would you define it in your own words?
 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.

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 Quote by frosty8688 No vertical asymptotes
This is wrong. Take the denominator and set equal to 0 to find the vertical asymptote(s).
 Vertical asymptote at x = 1.

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