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

Integrate

Integrate

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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.
 
The difference is between the [imath]9x[/imath] and [imath]9x^2[/imath] in the denominator.
For [imath]9x^2[/imath], regardless of the [imath]x<0[/imath] or [imath]x>0[/imath], it'll always return a positive value. Thus we only need to consider [imath]x\neq 0[/imath] for the horizontal asymptote. That's not the case for [imath]9x[/imath], 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)
1646154577227.png
20)
1646154639201.png
 
Ahhh, this really helps clear it up. Thank you. Such a great site.
 
Oh, but I take it that we are always to go both directions of infinity when trying to find horizontal asymptotes?
 
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