Sal finds the limits at positive and negative infinity of x/√(x²+1). Since the leading term is raised to an odd power (1), the limits at positive and negative infinity are different.Created by Sal Khan.
Let's say that f(x) is equal to x over the square root of x^2+1 and I want to think about the limit of f(x) as x approaches positive infinity and the limit of f(x) as x approaches neagative infinity. So let's think about what these are going to be. Well, once again, and I'm not doing this in an ultra-rigorous way but more in an intuitive way is to think about what this function approximately equals as we get larger and larger and larger x'es. This is the case if we're getting very positive x'es, in very positive infinity direction or very negative. Still the absolute value of those x'es are very very very large as we approach positive infinity or negative infinity. Well, in the numerator we only have only 1 term -- we have this x term -- -- but in the denominator we have 2 terms under the radical here. And as x gets larger and larger and larger either in the positive or the negative direction this x squared term is going to really dominate this one You can imagine, when x is 1 million, you're going to have a million squared plus one. The value of the denominator is going to be dictated by this x squared term. So this is going to be approximately equal to x over the square root of x squared. This term right over here, the 1 isn't going to matter so much when we get very very very large x'es. And this right over here -- x over the square root of x squared or x over the principle root of x squared -- -- this is going to be equal to x over -- -- if I square something and then take the principle root -- -- remember that the principle root is the positive square root of something -- -- then I'm essentially taking the absolute value of x. This is going to be equal to x over the absolute value of x for x approaches infinity or for x approaches negative infinity. So, another way to say this, another way to restate these limits, is as we approach infinity, this limit, we can restate it as the limit, this is going to be equal to the limit as x approaches infinity of x over the absolute value of x. Now, for positive x'es the absolute value of x is just going to be x. This is going to be x divided by x, so this is just going to be 1. Similarly, right over here, we take the limit as we go to negative infinity, this is going to be the limit of x over the absolute value of x as x approaches negative infinity. Remember, the only reason I was able to make this statement is that f(x) and this thing right over here become very very similar, you can kind of say converge to each other, as x gets very very very large or x gets very very very very negative. Now, for negative values of x the absolute value of x is going to be positive, x is obviously going to be negative and we're just going to get negative 1. And so using this, we can actually try to graph our function. So let's try to do that. So let's say, that is my y axis, this is my x axis, and we see that we have 2 horizontal asymptotes. We have 1 horizontal asymptote at y=1, so let's say this right over here is y=1, let me draw that line as dotted line, we're going to approach this thing, and then we have another horizontal asymptote at y=-1. So that might be right over there, y=-1. And if we want to plot at least 1 point we can think about what does f(0) equal. So, f(0) is going to be equal to 0 over the square root of 0+1, or 0 squared plus 1. Well that's all just going to be equal to zero. So we have this point, right over here, and we know that as x approaches infinity, we're approaching this blue, horizontal asymptote, so it might look something like this. Let me do it a little bit differently. There you go. I'll clean this up. So it might look something like this. That's not the color I wanted to use. So it might look something like that. We get closer and closer to that asymptote as x gets larger and larger and then like this -- we get closer and closer to this asymptote as x approaches negative infinity. I'm not drawing it so well. So that right over there is y=f(x). And you can verify this by taking a calculator, trying to plot more points or using some type of graphing calculator or something. But anyway, I just wanted to tackle another situation we're approaching infinity and or negative infinity and we're trying to determine the horizontal asymptotes. And remember, the key is just to say what terms dominate as x approaches positive infinity or negative infinity. To say, well, what is that function going to approach, and it's going to approach this horizontal asymptote in the positive direction and this one in the negative.