 In this video, I want to talk about taking limits as x approaches infinity or negative infinity with rational expressions, but this time involving radicals of some kind, particular emphasis will be put on square roots here, because they can add a little bit of difficulty for people. So consider the limit as x approaches infinity of x over the square root of x squared plus one. Now, in this one, it's very tempting to do what we did previously with rational functions taking their in behavior. And in some regard, we will be perfectly fine in doing that for which we can identify who is the dominant term on top the leading term that's just an excellent one term. And then in the denominator, we can see the dominant term is actually given the square root of x squared, which is fine. In which case, then we can think of this as the limit of x over the square root of x squared as x approaches infinity in this setting here, for which you can simplify this, you'll get the limit of x over x as x approaches infinity. And this will then give you the limit of one as x approaches infinity. This then agrees with what we might have anticipated. This function, its limit as x approaches infinity is going to equal one. Okay, now there is a little bit of a problem with this approach. And I'll kind of illustrate this by giving a comparison, right? If we were to do it how we originally considered this topic, take the limit as x approaches infinity of x over the square root of x squared plus one. If we wanted to clear out the leading terms, right, we times the top by one over x, the bottom will do the same thing one over x. In the numerator, it's pretty simple what's going to happen there, you're going to get the limit of one, that's what you would expect. But in the denominator, how do you multiply the one over x inside the square root? Well, the square roots kind of like an exclusive fancy club, you can't just go in in your street clothes, you got to go with a suit and tie, you have to look really elegant. So if you take one over x and want to distribute it through here, you're really going to have to write it as the square root of one over x squared. If you make that statement, it's like, okay, then the in the denominator, you're going to see the square root of x squared plus one times one over x squared, you distribute that through, you're going to get the limit of one over the square root of one plus one over x squared as x approaches infinity. And then we see that as x approaches infinity, one over x squared will go to zero. And so this limit then turns out to be one over the square root of one plus zero, that simplifies just to be one. So that's, that's an agreement, whether we saw it a moment ago. Okay, what's the concern there? What's the malfunction? Well, the malfunction comes into play mostly when you start taking limits of these square root functions, these these radical these rational square functions, as you go off towards negative infinity, because this can be a little of a worrisome issue. If you have x over the square root of x squared plus one, if you're like, oh, I'm just going to take the leading terms here, you're going to take the limit of x over the square root of x squared as x approaches infinity, negative infinity here, right? This is where the mistake I often see is at this moment, you're like, oh, what the square root of x squared is x. That's actually not true. When it comes to the square root of x squared, this is actually equal to the absolute value of x, for which if you're a positive number, you see no difference between the absolute value of x and x itself. But if you're a negative, because after all, you're approaching negative infinity, this thing is going to be negative. And therefore your denominator should actually be the absolute value of x, as x approaches negative infinity, for which when you take x divided by the absolute value of x, this actually gives you plus or minus one, dependent upon whether you're positive or negative, it's going to be positive one when x is positive. It's going to be negative when x is less than zero. It's undefined at zero, but as we're approaching negative infinity, we'll be nowhere near zero, so it's not a big deal. So it turns out that this is going to be negative one. If you're very careless on your simplification right here, then you might miss that there's a negative sign here. But how does it approach, how does it compare to this approach right here? If we try it that way, take the limit as x approaches negative infinity, x over the square root of x squared plus one. Same thing as before, we're going to take the top to be one over x. We're going to take the denominator to be one over x, like so, but we need to rewrite this so it goes inside the square root function. So whoops a daisy, we need to do the square root of one over x squared. But again, in this situation, the sign is incompatible. When you take the square root of something, that's inherently a positive object. So you have to stick out a negative so that this is actually so that this ratio in red is actually the number one. Because if you don't have the negative sign, the top is negative, the bottom is positive. And therefore, it's not actually, you didn't multiply by positive when you multiply by negative one. So you have to stick a negative sign in there. The consequence then gives you a negative sign in front of this whole thing, you're going to get one over the square root of one plus one over x squared, as x approaches negative infinity. And then you get the same thing again, negative one over the square root of one plus zero, you end up with negative one. I guess the thing I really want to caution you is whether you want to use this approach of flattening the function, or you kind of like this more simplified, just look at the dominant terms, that's okay, but you have to very much pay attention to the sign of the square root as you're approaching negative infinity. That sign can throw you off by a lot. You'll notice that this function, this function has a horizontal asymptote at x equal or y equals one, but it also has a horizontal asymptote at y equals one. This function has two different horizontal asymptotes. This is quite typical of rational expressions involving square roots. Let's take a look at another example here. Let's take the limit as x approaches infinity of the square root of two x squared plus one over three x minus five. We're going to attempt this one with my simplified approach. We're just going to look at the numerator's dominant term and the denominator's dominant term. We end up with the limit because as x goes to infinity, in the end, nothing really matters except for the leading term. You can get the square root of two x squared over three x, for which in the numerator, this will simplify as the square root of two times the absolute value of x over three x as x approaches infinity. Now, as x approaches infinity, positive infinity, we see that the absolute value of x is actually no, it's not distinguishable from x itself, so you get three x there. In which case then, oh, that's the worst drawn infinity in the history of the world there. The x will cancel out and we see in the end, this is actually a balanced rational function. We see that this limit turns out to be the square root of two over three. Notice here that we just looked at the coefficients because this thing was in fact balanced. It's kind of a curious thing there because you have an x on the bottom and on the top, you have the square root of x squared, which is essentially just x, so you see it's balanced here. But like I cautioned you in the previous example, when x approaches negative infinity, things are a little bit different, right? You get the square root of two x squared plus one over three x minus five. Like we did over here, you can look at just the dominant term and you're going to get the limit of just the square root of two x squared over three x. Dominance did not change in this situation. That does sometimes happen. For power functions, dominance does not change, but for exponentials, it gets a little bit more kooky. We'll do that in a different video. When you simplify this, you're going to get the square root of two times the absolute value of x over three x, as x approaches negative infinity. And this is the thing to keep track of since we're approaching negative infinity. The absolute value of x divided by x, in this case, is not positive one. It's equal to negative one because the absolute value is positive and x itself, though, is negative. So we're going to end up with the limit of negative square root of two over three as x approaches negative infinity. That is the limit will be negative, negative root two over three. So be very cautious about these type of calculations with, again, with these square roots and rational functions. Even when it's balanced, the right limit as you go towards positive infinity and the left limit as you go towards negative infinity, those could still be different in signs, like we saw in this one. The next example is also a little bit of a kooky thing. You'll notice that the previous examples, if we look at them, the initial the initial form of the limit here looks like infinity over infinity. That's why we can't just plug in infinity and hope for the best. In this example, if we plug in x equals infinity, we actually end up with infinity minus infinity. And so what is that? When you look at it, it's like, well, you have the square root of x squared subtracting x, you kind of hope that they're going to cancel each other out. And that's kind of what's happening here, but we have to be a little bit more careful. So when you have this infinity minus infinity, if you ever run across that, my strategy is somehow turn this into a fraction. We don't want infinity minus infinity, but we could live with infinity divided by infinity. And so because we have this square root expression, I'm going to rationalize the numerator. That is, I'm going to multiply the top by the square root of x squared minus one plus x, and then we do that to the denominator as well. We leave the denominator alone, just leave it factored, which there's really nothing to multiply out there, but you do need to multiply out the numerator. So we take the limit as x approaches infinity. In the numerator, when you forward that out, you're going to get x squared minus one, you're going to get the square root times x, you're going to get negative x times square root, those will cancel out, and then you're going to get negative x squared when you take x times x, and then you have the square root of x squared minus one plus x right here. In the numerator, you'll notice that the x squares cancel out, leaving you with a negative one in the numerator. In the denominator, you have the square root of x squared minus one plus x as x approaches infinity. If you check now, do we have the indeterminate form infinity minus infinity anymore? The answer is actually no, if we just plug in infinity, we're going to end up with negative one over the square root of infinity squared minus one plus infinity here. Simplifying this, when you take infinity square, it's infinity minus one is infinity, and when you take the square root of infinity, that's still infinity. So you get negative one over infinity plus infinity, which doubles up just to be infinity here, and when you divide by infinity, that's going to be zero in this situation. Since we have a negative on top, positive on the bottom, this would be, we're approaching zero from the right, excuse me, from the left, but again that level of specificity we don't necessarily need in this situation. And then to finish up this video, what if we do the other direction? What if we take the limit as x approaches negative infinity in that situation? Before we start rationalizing the numerator, I want to caution you here that we no longer have an indeterminate form. If we just plugged in infinity, or I guess I should say negative infinity into the situation, you're going to take negative infinity square minus one minus a negative infinity. When you square a negative, you're going to get positive, so this is going to be infinity squared minus one. But then we also here have a double negative, so this is going to be plus infinity. So we end up with infinity plus infinity, which then gives us infinity, right? It's just going to be infinity plus infinity, it's just infinity here. So this is again sort of a curious function, this ration, it's like a square root function, but we subtracted a line from it. On the right hand side, it has a horizontal asymptote at the x-axis, but on the left hand side, it actually goes off towards infinity. So in some regard, this graph kind of looks like an exponential function, although its decay won't be as steep as a typical exponential.