When and why do we consider a negative value for radicals when finding the horizontal asymptotes of limits at infinity. [Examples given]

[Integrate]

In 19.) We are shown that we are to consider a negative value for our limit at infinity. I can't seem to understand the given explanation in the solution.

In 20.) We are given an almost identical function however with higher powers. But this time it is not given a negative value for the solution. Why is this?


I'm sure it will be obvious in retrospect, but I can't seem to get there on my own.


Thank you in advance.

 

[BigBeachBanana]

The difference is between the $9x$ and $9x^2$ in the denominator.
For $9x^2$, regardless of the $x<0$ or $x>0$, it'll always return a positive value. Thus we only need to consider $x\neq 0$ for the horizontal asymptote. That's not the case for $9x$, thus we need to consider both cases. Furthermore, if you look at their graphs, 19) has 2 horizontal asymptotes, while 20) only has one.
19)

20)

 

[Integrate]

Ahhh, this really helps clear it up. Thank you. Such a great site.

 

[Integrate]

Oh, but I take it that we are always to go both directions of infinity when trying to find horizontal asymptotes?

 

[BigBeachBanana]

Oh, but I take it that we are always to go both directions of infinity when trying to find horizontal asymptotes?

Yes, that would be a good practice.