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All right, so this is a calculus problem in which they
want us to find the limit as x approaches infinity for a
given function as well as x approaching negative infinity
for that same given function.
And the function that we have is this
rational function here.
What they really want us to find with these two limits as
x approaches infinity and negative infinity are the
horizontal asymptotes.
All right?
So as we go off to a negative or positive infinity, we're
looking for horizontal asymptotes.
They want us to find all the asymptotes as well.
So we'll find the vertical ones also.
Let's see.
Let's find the horizontal asymptotes first.
All right?
Let's find the horizontal ones first.
To find the horizontal asymptotes, one way to look at
this, maybe probably the easiest way to look at this,
is to notice that the degree of numerator is a 4.
And the degree of the denominator is a 4.
So since the degree for the numerator is a 4--
let's see, the book I'm using says, since m is equal to n--
since the two degrees are the same, numerator and
denominator, then what we need to do is look at the leading
coefficients for both the numerator and the denominator
and simply just set our y equals-- this is going to be
our horizontal asymptote--
to whatever that ratio happens to be.
Well, obviously, the ratio or the number for a leading
coefficient up top is a 2.
And for the denominator, since I don't see one, it's an
automatic 1.
So, OK, so y equals 2 is the limit as
x approaches infinity.
Turns out, it's the same as x approaches negative infinity.
So that's one easy way to find the horizontal asymptotes.
In this case though, the book shows there is another method
that we could use if you wanted to.
Let me show you that as well.
OK?
When you're dealing with rational functions like these,
what you can do is you can concentrate on the denominator
and look at the denominator's largest exponent.
In this case, it's going to be a 4.
All right?
So I'm writing out my problem again here,
rewriting this thing.
I'm going to rewrite it just like this.
So noticing that the denominator's largest exponent
is a 4, what we can do with that x to the fourth is we can
divide every single term in the numerator and the
denominator by x to the fourth.
So I'm going to write it like this.
Divide everybody here by x to the fourth, even these guys on
the bottom.
So now let's say that the numerator's largest
exponent was a 5.
Doesn't matter.
You're concentrating on the denominator's, all of these
terms in the bottom, the
denominator's largest exponent.
That's the one you're looking for.
And then you're going to divide everybody by that same
largest exponent, in this case, x to the fourth.
What does that do for us?
Well here's what it does for us.
OK?
Here's what it does for us.
It says, all right, hey look.
These x to the fourths cancel out.
And so I'm simply left with a 2 right
there, coming from there.
OK?
Let's see. x to the third cancels out
with x to the fourth.
Well, it cancels out three of these x's down here.
So I'm left with 2 over x, because three of these cancel
out with three of these, leaving me with one more.
Now hey, look, two of these x's cancel out with two of
these x's on the bottom.
That leaves me with x squared on the bottom.
This is minus 60 over x squared.
And I'm going to do the same thing for these three terms in
the denominator.
So let's see.
This just going to be a 1 minus, let's see, this is 61
over x squared.
And out of this, nothing cancels.
I'm just going to leave that as 900 over x to the fourth.
Now the reason I'm doing that is because you should
recognize from calculus that as the limit approaches
infinity or negative infinity, these terms right here, like
this one, this one, this one, and this one, are all
going to go to 0.
So as we head off to both positive infinity and negative
infinity, as we head off to negative infinity as well,
these terms right here--
anything with an x or an x squared or an x to the third
or an x to the fourth, anything with an x in its
denominator--
these guys are all going to go to 0.
So do you see that we're left with just a 2 over 1?
And that's simply the case, no matter which direction we're
heading, positive or negative.
And so I'm just going to say, hey, the limit is 2.
No problem there.
All right, let's go a little bit further.
Now we have the horizontal asymptotes.
Let me write that up here.
So we have the horizontal asymptotes.
In this case, there's really only one
asymptote is y equals 2.
That's it.
OK, what about the vertical asymptotes for
this exact same problem?
Well, let's see.
To find the vertical asymptotes--
I'm going to go back to my original problem here--
I need to find out what value or values for x is going to
make this denominator a 0.
All right?
And it's not easy to see that right now.
But if I factor everything as much as I can up top and
factor these terms here in the bottom as much as I can, I
hope you see this is what I'll end up with.
All right?
I'll end up with--
these guys in the numerator all have a 2x squared in
common that I can take out.
I guess I'll show you the intermediary steps here.
I'm left with an x squared plus x minus 30.
And these three terms on the bottom turn into x squared
minus 36 and x squared minus 25.
OK, but it turns out that this trinomial in the numerator and
these two guys in the denominator can still be
factored further.
OK, so I'm going to factor these guys a little bit more.
I'm going to leave that 2x squared
alone in the numerator.
But this trinomial here up top, I can factor a little bit
more into x plus 6 and x minus 5.
And since both of these in the denominator are difference of
squares, I can do this.
All right, we can split up this x squared minus 36 as x
plus 6 and x minus 6.
And this x squared minus 25 is x plus 5 and x minus 5.
And now that I've factored as much as possible, can you see
that there are four values that would make
my denominator 0?
OK, there'd be a negative 6 coming out of this one,
positive 6 coming out of this one, negative 5 for that one,
and positive 5 for that one.
All four of those values would make my denominator a 0.
So the function is undefined at four of those points.
But not all four of them are vertical asymptotes.
Why not?
Well, do you see that this binomial and that one, since
they're the same on top and on the bottom, they cancel out,
as well as these.
So it turns out there are really only two vertical
asymptotes--
x equals 6 and x equals negative 5.
So there are really only two vertical asymptotes.
Now the last thing, if my math lab or your homework wants one
more thing out of this problem, and that is, OK, so
you have the two vertical asymptotes.
Fine.
But what is the function doing as far as, as we approach
these two values coming from either side?
So maybe the question is asking something like this.
Let me write our simplified form here.
I've got x minus 6 and x plus 5.
So this is the simplified form of our
original rational function.
And so we know--
here's 0 right there.
So we know that there is a vertical asymptote here at 6.
And we know that there's a vertical asymptote here at
negative 5.
Let's find out the behavior of the function.
What is the function going to do as we approach these
asymptotes, coming from either side, coming from either side?
Let's start off with this one.
If we approach negative 5 on the left side of negative 5,
as we come at it from the left side, the way I'm pointing
with my finger, all of these numbers
over here are negative.
Right?
Negative 6, negative 7, negative 8, et cetera.
So all these numbers here are negative.
That makes these two parentheses here, a negative
minus 6 stays a negative.
And a negative plus 5 stays a negative.
That means I've got two negatives in the denominator,
which is really just a positive since they're being
multiplied.
And a negative value squared stays positive.
So it turns out that everything over here coming
from the right hand side of the function, coming in this
direction here, everything is going to be positive.
Let's try out coming at negative 5 from the right hand
side of it, though.
Coming from the left, it's going to be positive.
Let's see what happens coming from the right.
Well let's see.
How about values of negative 4, negative 3, or negative 2,
or negative 1?
And you can pick any one you want.
How about negative 4 for instance?
If I plugged in a negative 4, negative 4 minus 6 is a
negative number.
Negative 4 plus 5, though, actually is a positive number.
So I'd have a negative times a positive, which is a negative.
So my denominator is negative.
My numerator is positive, because it's
being squared anyways.
Right?
So I've got a negative on the bottom and a positive on top.
Actually turns out, all of these values down here in
between negative 5 and 6 are negative numbers.
So coming at negative 5 from the right hand side is a
negative infinity, a negative infinity.
Coming at negative 5 from the left hand side
is a positive infinity.
Likewise, coming at 6 from the left hand side
is a negative infinity.
But coming at 6 from the right hand side-- let's see.
These numbers would be 7 or 8 or negative--
I'm sorry, these are all positives.
Right?
Positive 8, positive 9.
You can plug in any number you want over here.
Let's say 8.
If we plugged in 8, 8 minus 6 is positive.
And 8 plus 5 is positive.
And my numerator's automatically positive.
So the whole thing is positive.
So we're talking about a positive infinity, a positive
infinity that's going on on this side coming at 6 from the
right hand side.
So you could say, all right, that we have vertical
asymptotes.
As far as the vertical asymptotes are concerned, you
could say that, look, the limit, as we approach negative
5 coming from the left hand side, is going to be, for this
particular function, is going to be a positive infinity.
That's this thing here.
And, likewise, the limit, as we approach--
you can see that or not--
the limit, as we approach negative 5 coming from the
right hand side of this function,
is a negative infinity.
When you approach negative 5 coming from the right hand
side is a negative infinity.
And I could say that the limit as we approach 6 coming from
the left hand side, 6 coming from the left hand side here,
is this negative infinity for function.
And the limit as we approach 6 coming from the right hand
side of our function, here we go, coming at 6 from the right
hand side is a positive infinity as we have up there.
So there you go.
I hope that helps.