# Thread: Which of these horizontal asymptote equations are wrong?

1. ## Which of these horizontal asymptote equations are wrong?

Hi, I have to match 6 functions with the equation for its horizontal asymptote, but I can't get it right.

I know at least 1 or more are wrong, can anyone tell me which, and how to solve it/them?

1. f(x) = (6x^3 + 2x + 1) / (2x^3 - 10x^2 + 13x +100)
y = 3

2. f(x) = arccot(x^2 - x^4)
y = pi

3. f(x) = arccot(x)
y = 0 and y = pi

4. f(x) = e^x
y = 0

5. f(x) = arctan(x)
y = pi/2 and y = -pi/2

6. f(x) = (6x^2 + 2x + 1) / (2x^3 - 10x^2 +13x +100)
none

Thanks!

2. To find horizontal asymptote you take the limit to infinity.

I'll the first 1 as an example

$\displaystyle \lim_{x \to \infty} \frac{6x^3 + 2x + 1}{2x^3 -10x^2 + 3x +100}$

now divide by the largest degree of the polynomial in the denominator

$\displaystyle \lim_{x \to \infty} \frac{6+ \frac{2}{x^2} +\frac{1}{x^3}}{2 - \frac{10}{x} + \frac{3}{x^2} + \frac{100}{x^3}}$

so you end up with 3 when taking the limit to infinity.

You could also use L'Hopital rule

3. Originally Posted by 11rdc11
To find horizontal asymptote you take the limit to infinity.
and negative infinity, as there can be as many as two horizontal asymptotes

4. Yes I know that, but my question is which one of those functions doesn't match the equation for its horizontal asymptote i've given?

5. Originally Posted by coldfire
Hi, I have to match 6 functions with the equation for its horizontal asymptote, but I can't get it right.

I know at least 1 or more are wrong, can anyone tell me which, and how to solve it/them?

1. f(x) = (6x^3 + 2x + 1) / (2x^3 - 10x^2 + 13x +100)
y = 3
correct

2. f(x) = arccot(x^2 - x^4)
y = pi
incorrect

3. f(x) = arccot(x)
y = 0 and y = pi
incorrect

4. f(x) = e^x
y = 0
correct

5. f(x) = arctan(x)
y = pi/2 and y = -pi/2
correct

6. f(x) = (6x^2 + 2x + 1) / (2x^3 - 10x^2 +13x +100)
none
incorrect

we already told you how to solve them

6. Thanks for the reply.

Can you show me a way to solve the ones I got wrong (specifically)?

7. Originally Posted by Jhevon
and negative infinity, as there can be as many as two horizontal asymptotes
Thanks I knew I forgot something lol

8. $\displaystyle f(x) = \frac{6x^2 + 2x + 1} {2x^3 - 10x^2 +13x +100}$

Ok using L'hopital rule

$\displaystyle \lim_{x \to \infty} \frac{6x^2 + 2x + 1} {2x^3 - 10x^2 +13x +100}$

$\displaystyle \lim_{x \to \infty} \frac{12x+2}{6x^2 -20x + 13}$

$\displaystyle \lim_{x \to \infty} \frac{12}{12x -20}$

so you have an asymptote at 0

Edit: o yea remember to test negative infinity aswell

9. thanks guys!