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 gonna do is we're gonna look at the rates of growth as x goes 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 gonna go out to infinity. So we're gonna have let f(x) and g(x) be positive for x sufficiently large so we're gonna have one of two things that are gonna occur if f grows faster than g as x goes to infinity we know that the limit as x goes to infinity of f(x) over g(x) would have to go infinity so if f is getting bigger our y-value are getting bigger than our g values as we're going out towards infinity for x that's gonna go to infinity. A different way of saying that is the limit as x goes this to infinity of g(x) divided by f(x) is gonna go to zero this may also be thought of as g is growing slower than f as x goes to infinity Our other cases if they're growing at the same rate and if the going at the same rate then 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 L'Hopital'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 gonna write a limit and we're gonna put that limit in fraction form and we're gonna realize that if I stick in infinity for x I'm gonna get infinity over infinity which is definitely an indeterminate form we're gonna use L'Hopital's rule so we're gonna take the derivative of the top which is gonna turn into 40x cubed plus 30 over the derivative 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 gonna get 120x squared over e to the x when we take the derivative of top and derivative of bottom indeterminate form again so 240x over e to the x these are all as limit goes to infinity, x goes to infinity. Indeterminate again so one more time of L'Hopital's rule limit a x goes to infinity of 240 over e to the x. Well now 240 is a constant it's no longer an indeterminate form so it's zero divided by infinity so we know that equals zero 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 gonna be going much faster to a higher y-value the next example we want to compare xln(x) minus x once again to e to the x well we're in indeterminate form if we'd stuck in infinity here we would've gotten infinity over infinity if we'd factored out an x we would've had ln(x) minus 1 and ln(x) is infinity minus 1 times infinity so we're gonna do L'Hopital's rule we're gonna use product rule in the beginning the derivative of x is just 1 so we get ln(x) plus the derivative of ln(x) is one over x times that original x and the derivative of this negative x down here is negative one all over e to the x. If we simplify it up these last two terms are gonna to cancel and we're gonna get ln(x) over e to the x putting it as x goes infinity we also get an indeterminate form so we're gonna use L'Hopital's rule again If we use L'Hopital's rule again we get: the limit as x goes to infinity of one over x over e to the x. While we could rewrite that simplifying the complex fraction into one over xe to the x as x is going to infinity we can see that goes to zero so xln(x) minus x grows slower than e to the x so the next example square root of one plus x to the fourth and we're gonna compare that again be e to the x so the limit as x goes to infinity, this is indeterminate form we're actually gonna take the square root of it all first, 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 one plus x to the fourth over e to the 2x. If we did L'Hopital's rule repeatedly we'd end up eventually realizing that the top's gonna turn into a constant just like in the first example but the bottom still gonna have e to the 2x times some constant so we know that this is gonna go to zero so square root of one plus x to the fourth grows slower than e to the x what about if we had xe to the x? Sometimes we don't have to use L'Hopital's rule, sometimes we can simplify algebraically first so the e to the x's 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 minus one over e to the x. In this case, we're gonna simplify it algebraically first and if we had the bases the same and we're dividing, we're gonna subtract the exponents so x minus one minus x means the x's will cancel and we get e to the negative one. Well, limit as x goes to infinity of e to the negative one. There is no x here so it's just gonna be one over e thus e to the x minus one 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(x) divided by g(x) goes to zero this is notated as f is a 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 o f is of, at most, the order of g as x goes to infinity if there is a positive integer M for which there ratio is less than or equal to M for x sufficiently large enough this case f is big O of g f little o of g implies f big O of g, for positive functions with x large enough so if f and g are growing at the same rate then f of big G, big O of g and g of big O of f have to be true let's look at an example. We want to know if one divided by x plus three equal to big O of one over x. So we're gonna do the limit as x goes infinity of one over x plus three divided by one over x simplifying that complex fraction up algebraically, we get limit as x goes to infinity of x over x plus three using L'Hopital's rule, we'd get one over one which would equal one, 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 two, one is less than or equal to two hence true let's look at another example we have one over x minus, oh let's go back, oh nope right there. One over x minus one over x squared is that little o of one over x, so we're gonna do the ratio again this time I'm gonna do it and algebraically but I'm gonna multiply by x squared over x squared, just to make it a little easier I'd get x minus one all over x using L'Hopital's rule we'd have 1/1 which would equal one but the little o's stated that it had to go zero. The fact that it's not going to zero means that for the little o, it had to be a false statement so that one's false the last example we wanna know how these two are gonna relate, and so when we look at square root of x plus four plus x and square root of x four minus x cubed. We wanna know how these limits are gonna relate, so we're just gonna choose some function that we do know and we're gonna 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 cubed because the x to the fourth is the stronger of this polynomial the square root of an x to the fourth would be x squared, so let's compare both of these terms or both of these functions that x squared. So if I had the limit as x goes to infinity of the square root of x to the four plus x over x squared to start with I could think about x squared as the square root of x to the fourth simplifying this down, I'd get one plus one over x cubed. Now when x goes out to infinity one over something really really big a zero so we'd get the square root of one or one so, the square to root of x to the four plus x and x squared are gonna grow at the same rate the square root of x to the fourth plus x equals the big O of x squared and x squared 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 cubed over x squared. Gonna do the same concepts square root make that square root of x to the fourth instead of x squared and simplify. So the limit as x goes to infinity of one minus one over x the square root of that whole thing, one over something really big a zero so we're gonna get one thus this new function and x squared also grow at the same rate if we know that two functions grow at the same rate as a third function, we know that the two functions have to grow at the same rate this one is 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