Hello, wonderful mathematics people. This is Anna Cox from Kellogg Community College. We frequently want to figure out what's happening to functions. And So what we're going to do is we're going to look at the rates of growth as X goes to Infinity, and we're going to compare them to functions that we already know. Like if we thought about E to the X, we know that E to the X when X gets big enough is going to go out to Infinity. So we're going to have let F of X&G of X be positive for X sufficiently large. So we're going to have one of two things that are going to occur. If F grows faster than G as X goes to Infinity, we know that the limit as X goes to Infinity of F of X / g of X would have to go to Infinity. So if F is getting bigger, our Y values are getting bigger than our G values as we're going out towards Infinity for X, that's going to go to Infinity. A different way of saying that is the limit as X goes to Infinity of G of X / F of X is going to go to 0. So this may also be thought of as G is growing slower than F as X goes to Infinity are other cases if they're growing at the same rate and if they're going at the same rate than the limit of that ratio would have to equal L where L is a positive constant. And what we're going to do is we're going to use Lopa Tall's rule frequently if we're in an indeterminate form. So our first example is asking us to use 10X to the fourth plus 30X plus one and compare it to E to the X. So we're going to write a limit, and we're going to put that limit in fraction form. And we're going to realize that if I stick in Infinity for X, I'm going to get Infinity over Infinity, which is definitely an indeterminate form. We're going to use Lopita's rules. So we're going to take the derivative of the top, which is going to turn into forty X ^3 + 30 over the derivative at the bottom, which is E to the X. Well, if we look at putting in Infinity again, we're also in an indeterminate form. So we're going to get one twenty X ^2 / e to the X when we take the derivative top and derivative bottom indeterminate form again. So 240 over X240X over E to the X. These are all as limit goes to Infinity, X goes to Infinity indeterminate again. So one more time of Lo Patel's rule limit as X goes to Infinity of 240 / e to the X. Well, now 240 is a constant, it's no longer an indeterminate form, so it's 0 divided by Infinity. So we know that equals 0. Thus the 10X to the fourth plus 30X plus one grows slower than E to the X when X is out at Infinity. If we thought about looking at this graphically, 10X to the fourth would be the strongest term in this polynomial. So it would sort of look like a quadratic or a fourth degree. But E to the X is going to be going much faster to a higher Y value. The next example, we want to compare XL N X -, X once again to EX. Well, we're in indeterminate form. If we had stuck in Infinity here, we would have gotten Infinity over Infinity. If we had factored out an X, we would have had LNX -1, and LNX is Infinity -1 times Infinity. So we're going to do Lopez rule. We're going to use product row in the beginning. The derivative of X is just one. So we get L&X plus. The derivative of L&X is 1 / X times that original X. And the derivative of this negative X down here is -1 all over E to the X. If we simplify it up, these last two terms are going to cancel, and we're going to get L and X / e to the X putting in it as X goes to Infinity. We also get an indeterminate form. So we're going to use Lopital's rule again. If we use Lopital's rule again, we get the limit as X goes to Infinity of 1 / X / e to the X. Well, we could rewrite that, simplifying that complex fraction into one over xe to the X. As X is going to Infinity, we can see that goes to 0, so XL and X -, X grows slower than E to the X. So the next example, sqrt 1 plus X to the 4th, and we're going to compare that again to E to the X. So the limit as X goes to Infinity. This is indeterminate form. We're actually going to take the square root of it off 1st and the square root can move in and outside the limit because the square root doesn't have an X in it, the actual square root function. So here, limit as X goes to Infinity of 1 + X to the 4th over E to the 2X. If we did Lopa Tau's rule repeatedly, we'd end up eventually realizing that the top's going to turn into a constant, just like in the first example, but the bottom's still going to have E to the 2X times some constant, so we know that this is going to go to 0. So sqrt 1 + X to the 4th grows slower than E to the X. What about if we had xe to the X? Sometimes we don't have to use Lopital's rule. Sometimes we can simplify algebraically first. So the E to the XS here would cancel and we'd have limit as X goes to Infinity of X, which is Infinity. So xe to the X grows faster than E to the X. Let's look at one more limit as X goes to Infinity of E to the X -, 1 / e to the X. In this case, we're going to simplify it algebraically first. And if we had the basis the same and we're dividing, we're going to subtract the exponents. So X -, 1 -, X means the X's will cancel and we get E to the -1. Well, limit as X goes to Infinity of E to the -1. There is no X here, so it's just going to be 1 / e. Thus E to the X -, 1 grows at the same rate as E to the X. A function F is of smaller order than G as X goes to Infinity. If the limit of F of X / g of X goes to 0, this is notated as F is little O of G. That's a little letter O. Another way of saying F grows slower than G as X goes to Infinity. F is little O of G, so F grows slower. Little OF is of at most the order of G as X goes to Infinity if there is a positive integer M for which the ratio is less than or equal to M for X sufficiently large enough. This case F is big O of GF. Little O of G implies F big O of G for positive functions with X large enough. So if F&G are growing at the same rate, then F of big G, big O of G&G of big O of F have to be true. Let's look at an example we want to know. Is 1 / X + 3 equal to big O of 1 / X? So we're going to do the limit as X goes to Infinity of 1 / X + 3 / 1 / X. Simplifying that complex fraction up algebraically, we get limit as X goes to Infinity of X / X + 3. Using Lopita's rule, we'd get 1 / 1, which would equal 1. So this one is true because we have the ratio less than or equal to some M There's a positive integer M We don't know what the integer M is, but there is one that we can say. So in this case, if we said M was 21 is less than or equal to two, hence true. Let's look at another example. We'd have 1 / X minus. Oh, let's go back. Oh, Nope, right there, 1 / X -, 1 / X ^2 is that little O of 1 / X So we're going to do the ratio again. This time I'm going to do it algebraically, but I'm going to multiply by X ^2 / X ^2 just to make it a little easier. I'd get X -, 1 all over X using Lopital's rule. We'd have 1 / 1, which would equal 1, but the little O stated that it had to go to zero. The fact that it's not going to 0 means that for the little O it had to be a false statement. So that one's false. The last example, we want to know how these two are going to relate. And so when we look at sqrt X + 4 + X and square root of X 4 -, X ^3, we want to know how these limits are going to relate. So we're just going to choose some function that we do know, and we're going to do each one separately and then see by our transitive property how the two would relate. Well, if we thought about ignoring the plus X and the minus X ^3 because the X to the 4th is the stronger of this polynomial, the square root of an X to the fourth would be X ^2. So let's compare both of these terms, or both of these functions to X ^2. So if I had the limit as X goes to Infinity of the square root of X to the fourth plus X / X ^2 to start with, I could think of that X ^2 as the square root of X to the 4th. Simplifying this down, I'd get 1 + 1 / X ^3. Now when X goes out to Infinity, one over something really really big is 0, so we'd get sqrt 1 or one. So the square root of X to the fourth plus X and X ^2 are going to grow at the same rate. The square root of X to the fourth plus X equals the big O of X ^2, and X ^2 equals big O of square root of X to the fourth plus X. Now let's look at the second function and see what happens. So the limit as X goes to Infinity of the square root of X to the fourth minus X ^3 / X ^2 going to do the same concept square root. Make that square root of X to the 4th instead of X ^2 and simplify. So the limit as X goes to Infinity of 1 -, 1 / X the square root of that whole thing one over something really big as zero. So we're going to get one. Thus, this new function and X ^2 also grow at the same rate. If we know that two functions grew at the same rate as a third function, we know that the two functions have to grow at the same rate. This one's really important because basically it says that if I have X to some power as I'm going out to Infinity, anything that's added or subtracted to that is really pretty much irrelevant. We're only looking at the highest term, the highest degree. Thank you and have a wonderful day.