### Video Transcript

Find the horizontal and vertical
asymptotes of the function π of π₯ equals three π₯ minus sin π₯.

We recall the horizontal line π¦
equals πΏ is an asymptote to the function π¦ equals π of π₯ if the limit as π₯
approaches either positive or negative β of π of π₯ is equal to πΏ. Then, for a vertical asymptote, we
say π₯ equals π is an asymptote if the limit as π₯ approaches π from the left of
π of π₯ is either positive or negative β. If the limit as π₯ approaches π
from the right of π of π₯ is either positive or negative β. Or if the limit as π₯ approaches π
of π of π₯ is either positive or negative β.

Now, in this question, π of π₯ is
given by three π₯ minus sin π₯. So letβs begin by finding the limit
as π₯ approaches β of three π₯ minus sin π₯. Then, we recall that the limit of
the sum or difference of two functions is equal to the sum or difference of the
limits of those respective functions. And we can write this as the limit
as π₯ approaches β of three π₯ minus the limit as π₯ approaches β of sin π₯. As π₯ approaches β, three π₯ itself
also approaches β whereas sin π₯ can only take values in the closed interval from
negative one to one. And so the limit as π₯ approaches β
of three π₯ minus sin π₯ is β. This isnβt a constant value as
required of πΏ. So thereβs no horizontal asymptote
here.

We should check the limit as π₯
approaches negative β. Once again, we split it up into the
limit of three π₯ minus the limit of sin π₯. As π₯ approaches negative β, three
π₯ itself also approaches negative β. Once again though, sin π₯
oscillates between negative one and one. This has very little impact on a
number as large as negative β. So the limit as π₯ approaches
negative β of three π₯ minus sin π₯ is negative β. And we can say that there are no
horizontal asymptotes.

Now, weβll consider the vertical
asymptotes. Weβll split our limit up. And weβll look for the limit as π₯
approaches π of three π₯ minus the limit as π₯ approaches π of sin π₯. We want to find a situation where
this might be equal to either positive or negative β. Well, we saw that the only way for
the limit of three π₯ to be β is if π₯ itself approaches β. Similarly, for three π₯ to approach
negative β, π₯ itself must approach negative β whereas the range of sin π₯ is the
closed interval from negative one to one. So the only way for the limit of
our function to approach positive or negative β is if π₯ itself approaches positive
or negative β. So actually, there are no vertical
asymptotes either. And so the function π of π₯ equals
three π₯ minus sin π₯ has no horizontal or vertical asymptotes.