00:01
All right, so this is quite a tricky problem.
00:03
So we're going to look at the function, f of x is equal to x minus 1 over x of the 2 thirds minus 1.
00:09
And we're particularly interested with discussing any asymptotes that it might have, right? so we're going to try and find the end behavior of the function.
00:16
So what happens is we go to infinity or minus infinity, and then any potential vertical asymptotes that there are.
00:21
So let's try and figure out the horizontal asymptotes first and foremost, right? so to do that, we want to find a limit as x goes.
00:30
To infinity of f of x and we want to find the limit as x goes to minus infinity of f of x right so in order to do this we're going to try and do some algebraic simplification or actually just rearranging a little bit of this function to be able to solve it more more quickly so those of you who maybe have seen some other results in limits like lopetal's rule you'll know this answer immediately but if you haven't gotten there yet totally fine we'll just going to use basic algebra here so what i'm going to do is i'm going to start off by writing the function f of x i'm going to try and factor out a power of x, and in this particular case, the power of x i'm going to factor out of the numerator and the denominator is actually x to the two -thirds.
01:12
And i'll show you why in just one second.
01:14
So if i factor out an x to the two -thirds from the top, what's left over? well, if i pull out an x to the two -thirds from x, i have an x to the one -third left.
01:23
And if i pull out an x -to -the -thirds from one, i get a minus one over x to the two -thirds.
01:31
So that's how i factor the numerator.
01:33
And if i do the same thing on the denominator, pulling out an x of the two thirds, i'll have a one left over from my first term, and i'll have a minus 1 over x of the two thirds from my second term.
01:44
Okay.
01:45
And in the usual fashion, what this is going to allow us to do is cancel off this factor that we've pulled out.
01:49
And so now we've got an alternate representation for f of x.
01:52
We've got the f of x is equal to x to the one third minus 1 over x to the two thirds over over 1 minus 1 over x of 2 thirds.
02:00
So why did i pull out x to the 2 thirds? well, because now we can see that in any limit as x is getting bigger in magnitude, so as x is going to minus infinity or deposit infinity, these two terms right here go to zero, right? these two terms, these red terms of unboxing, these two terms are going to go to zero.
02:19
And this term right here is going to be the only thing that actually ends up determining this limit, right? so we only have one term lift over that's going to determine the value of this limit, and all the other ones that involve x are going to go to zero.
02:30
And of course that one's going to be there.
02:31
This one's still going to be there, but dividing by one is just one, or it's just the original value.
02:35
So it's not going to actually end up changing anything.
02:37
So let's look at each limit individually.
02:39
Let's look at the limit as x goes to infinity, the positive one first.
02:43
What happens to the function as we go to infinity? let's write it in our new form, right? so our new form is x to the one -third minus one over x to the two -thirds over one minus one over x to the two -thirds.
02:56
Well, like i said a second ago, these two terms, the second terms, these both go away, right? these both go to zero.
03:03
Those both go to zero.
03:04
And so this limit is approaching, you know, it's getting, it's getting closer and closer to just the limit of, as x goes to infinity, of just x the one -third, right? so what happens to x the one -third as we go to infinity? well, that's pretty easy to see.
03:17
It just goes to infinity, right? it grows slower than x, but it still goes to infinity.
03:21
It's still getting bigger.
03:22
And the same exact thing is going to be true if we think the limit as x goes to minus infinity, right? so we can skip the other previous steps in algebra, because they're all the exact same.
03:30
If we go to minus infinity now, it's, it's an odd reciprocal exponent, right? so it's one -third, not one -half or anything like that.
03:38
So this is totally fine, and the sign is actually preserved, right? when we take the cube root of a negative, we get a negative back out.
03:44
And so this is going to go to minus and then.
03:46
All right, so this is the relatively straightforward part of this problem.
03:50
Now, this is, well, in summary, before i get there, in summary, this tells us that we have no horizontal asymptotes, right? because as we go really far out to the right, it's blowing up.
04:00
And as we go really far out to the left, it's blowing up.
04:02
We can't have any horizontal asymptotes.
04:04
That takes care of that.
04:05
No horizontal asymptotes.
04:07
We should write that in.
04:08
So no horizontal asymptotes.
04:14
Okay? but what about the vertical asymptotes? and this is where the problem gets a little tricky.
04:18
So let's look at the function again.
04:20
So we have f of x is equal to x minus 1 over x of 2 thirds minus 1.
04:25
Now if you look at this for a second, if you just glanced at it, you might think there's a vertical asymptote at x is equal to 1, right? because if you plug in one into the denominator, it looks like it's zero, right? and it is.
04:38
It is zero, right? but the thing is the numerator also goes to zero.
04:42
And so our question is, does it go to zero at a rate? do they go to zero at a rate that actually makes it end up having a limit? and turns out in this particular problem, they do.
04:51
So there is actually no vertical asymptote at one.
04:54
So how do you determine that? because i've just told you that, but how do you see that? and this is where the algebra gets a little bit tricky.
05:01
So let's start with the denominator because we're going to just rearrange this function again, just like we did in the first part.
05:06
We're going to rearrange this in a slightly easier way to see it.
05:10
So what's something that we could do with the denominator? it's got a reciprocal exponent, so that's kind of hard to see.
05:19
But if you look at this a little bit in a slightly different way, if i were to just write this as x to the one -third all squared minus one.
05:28
And if i even put a square on the one, well, now i can see that the denominator is actually secretly a difference of two squares.
05:36
What are the two squares that it's a difference of? well, it's the square of one, x to the one -third, and it's the square of one.
05:42
So we can actually factor that, using our difference of two squares formula, we can factor that as x to the one -third minus one, times x to the one -third plus one, right? this is a very counterintuitive thing to see.
05:56
Right it's not not natural um but we can do it and so why does this help us well because i mean the function looks a bit more complicated now and the numerator still doesn't cancel with anything so how does that help us well we're going to actually do something that's not really that standard of a thing to teach in an algebra course but it's an undeniably useful tool which is the equivalent of differences of two squares with cubics right so there's actually a difference of two cubes formula as well now there's one for plus and minus but we're we're going to only care about the formula for the difference of two cubes.
06:31
So a cubed minus b cubed.
06:33
So it turns out, i'm not going to show you how this is true, but it turns out that you can actually factor this always as a minus b times a squared plus ab plus b squared.
06:45
This is an immensely useful formula.
06:47
And in this particular problem, it's necessary.
06:49
We need it.
06:50
And we're going to do that to the numerator because i can secretly write the numerator.
06:54
I could secretly write x minus 1 as x to the 1 third to the power of 3 minus 1 to the power of 3.
07:04
So it's actually secretly a difference of 2 cubes as well.
07:07
And in doing that, what i see is that my f of x actually can be written as x to the 1 3rd minus 1.
07:16
Okay, that's my a minus b.
07:18
And then everything that's left over is going to be an x to the 2 thirds, right? that's my a squared plus x to the one -third times one, right? that's my a -b term.
07:31
I'm going to drop that one, if you guys don't mind.
07:33
And then my last term is going to just be plus one, right? that's my b -squared term.
07:38
And so now i have f -of -x as equal to this nasty -looking thing over x -to -one -third minus one times x to the one -third plus one, right? hopefully this is okay so far.
07:51
And what i see now is that i have a common factor in both the numerator and the denominator.
07:56
And not only do i have a common factor, but the factor that i'm canceling out is the thing that was letting the denominator go to zero in the first place when we're approaching x is equal to 1.
08:05
So there is actually no vertical asymptote at x is equal to 1.
08:09
So there's no vertical asymptote, even though it seemed like it from the beginning, there's no vertical asymptote at x is equal to 1.
08:18
That doesn't happen because that factor actually ends up canceling out.
08:21
But there is still going to be vertical asymptote.
08:24
This is the crazy part about this problem.
08:25
So let's write down the function now that we've canceled out that factor.
08:28
The function is now x to the two -thirds plus x to the one -third plus one, all over x to the one -third plus one.
08:39
Okay.
08:40
So what is the vertical asymptote going to be here? well, the vertical asymptote is going to, again, just be the thing that makes the denominator go to zero.
08:49
So what makes the denominator go to zero? well, let's see.
08:52
Let's set this equal to zero.
08:53
Let's set this thing equal to zero.
08:55
We have x to the one third plus one equals zero.
08:59
Solving gives us x to the one third equals minus one, which looks illegal, but it's totally fine.
09:06
Taking both, if we raise both sides to the power of three, that gives us x is equal to minus one.
09:10
And it turns out there actually is in fact an asymptote of x is equal to minus one.
09:15
So technically here, you need to verify that this isn't going to also cancel with something in the numerator.
09:20
But you can do that by just plugging in minus one into the numerator.
09:23
So if we plug minus one into the numerator, we get minus one to the two thirds plus minus one to the one third, plus one, okay? minus one to the two thirds is one.
09:35
Minus one to the one third is minus one, and one is one.
09:39
So this is actually all equal to one here...