so in the last video we looked at the limit as X goes to Infinity of 1 over X we sent our X's to infinity and we saw that one over X went to zero but the same is true for other related functions like instead of one over X we could have 1 over X squared or 1 over the square root of x and similar things happen right as X goes to Infinity the bottom here is still going to get infinitely large which is going to make that ratio 1 over that value get infinitely small same thing here as X goes to Infinity the square root of x is still going to get bigger and bigger and bigger making this overall value smaller and smaller and smaller but that is sort of an intuitive approach if you want to go ahead and plot some points we can so if x is 10 squared is 100 and all of a sudden we have one hundredth if x was equal to a 100 a hundred squared is 10 000 and then all of a sudden you have ten thousandth and so what happens is when you take a big number and then you square it gets even bigger and so for therefore these numbers are actually even smaller than these numbers or they are getting smaller more quickly and so these numbers are definitely going to zero as these values are going off to Infinity now with one over the square root of x the values still go to zero they just don't go to zero as quickly so here if x was a hundred plugged 100 in here the square root of 100 is 10 and we get one tenth if x was ten thousand the square root of ten thousand is a hundred and we end up with 1 100th if x was a million the square root of a million is a thousand and we end up with one one thousandth and so again these values are still approaching zero they're just not approaching zero as quickly as in the previous two examples so there's this theorem says if R is a rational number positive rational number then the limit as X goes to Infinity of one over x to the r equals zero so whether this is a square like a 2 or a cube like a 3 or a square root where R is equal to one half they are all heading to zero as long as R is a positive number further if R is such that x to the r is defined for all X then we can also send X off to negative infinity and that limit is also zero and that mostly has to do with like even Roots you cannot take the square root of a negative number however you could take the cube root of a negative number and so if R is such a power that uh it's defined for all X's sorry so there's that x to the r is defined for all X's then this limit also holds true so we can see what that looks like in just a few quick examples here so as X goes to Infinity 1 over x to the three halves is zero right as X goes to Infinity this bottom is getting infinitely large so therefore 1 over that value should be getting smaller and smaller and smaller next one if I say let's take the limit as X goes to Infinity over one over x to the two thirds the same thing happens as X goes to infinity x to the two thirds gets infinitely large so 1 over that value should get infinitely small and again approach zero but let's recall our rational exponent notation x to the two-thirds is the same thing as the cube root of x squared or x squared to the cube root but since I am taking in either case an odd root you can take odd roots of negative numbers and so I can actually plug in negative values so in this case I could also send X off to negative Infinity in which case this thing would still approach zero okay let's take a look at a rational function let's send the limit as X goes to Infinity of 5x squared plus 2X minus 3 over 3x squared plus 9x now what you may start to consider is as X gets really really big this is going to get really big this is going to get really big we're adding them together all this stuff on the bottom is going to head off to Infinity but then as X gets big this gets big this gets big that taking away 3 is not going to do very much again this top is going to go off to positive Infinity so the numerator is getting infinitely large and the denominator is getting infinitely large and so where is this overall limit going and that's where we have some work to do and so one of the things I mentioned is in this section we're also going to consider some algebraic methods and so one common method is to rearrange this and I would do that by multiplying top and bottom by 1 over x squared and if I were to distribute 1 over x squared to the top and 1 over x squared to the bottom we would come up with an equivalent ratio we're now when I multiply 1 over x squared times this 5x squared that is just 5 plus 1 over x squared times this that is just 2x over x squared that is 2 over X and times this that would be negative 3 over x squared that is the numerator now the denominator 1 over x squared times 3x squared should be 3 1 over x squared times 9x would be plus 9 over X now oftentimes in the textbooks if you look at a problem like this they then break this expression up into using all the different limit properties like the limit of a quotient is the quotient of the limits the limit of a sum is the sum of the limits but we're going to use sort of the more intuitive approach when we take a look at some of these so now X is going to Infinity so as X gets infinitely large this value gets infinitely small and goes to zero right two over ten two over a hundred two over a thousand they're getting smaller and smaller and smaller this value is getting closer and closer and closer to zero this value is getting closer and closer and closer to zero so the numerator has a 5 plus a whole bunch of values which are getting closer and closer and closer to zero so this numerator is approaching 5. the denominator has a 3 plus lots of values that are approaching zero so the denominator is approaching 3. so this limit should be approaching Five Thirds it's mixing in a little bit of algebra to turn it into something where we can now see where the limit's going and then using our intuitive methods to figure out where it should be heading all right now again for this one let's send X off to negative Infinity well as X goes to negative Infinity again this quantity should be getting infinitely large but negative this quantity should be getting infinitely large but negative but that doesn't tell us where the limit's heading so let's algebraically rewrite this in a different way so we can actually see where it's headed so in the case in this one I'm going to multiply numerator and denominator by 1 over X now when I do that x times 1 over X is one minus 1 over x times two is minus two over X 1 over x times x is one plus one over x times three should be three over X and now I want to send X off to negative Infinity we can go ahead and do that as X gets infinitely large but negative this value still is approaching zero and as X gets infinitely large with negative values this is still approaching zero which means the overall top should be approaching 1 and the overall bottom should be approaching one so this should be approaching one over one or just one all right let's do one more of these so now there's a difference here on this first one we had a rational function where we had a 5x squared and a 3x squared on the second one we had a rational function where we had a 1X and a 1X but the degree of the numerator and the denominator were the same and this one we've got a rational function but the degree of the numerator and the denominator are not the same the numerator has a first degree polynomial the denominator is a second degree polynomial and so you have two options here I will try to highlight them this was an X so I multiplied top and bottom by 1 over X this one was an x squared so I multiplied top and bottom by 1 over x squared however for this next problem we have an X here and an x squared and so you can actually multiply top and bottom by 1 over X or you can multiply top and bottom by 1 over x squared and so I've gone ahead and done this both ways so first one multiply top and bottom by 1 over X 1 over x times 2x 1 over x times one we should get two plus one over X 1 over x times 3x squared should be 3x 1 over x times nine X should be 9 and 1 over x times one should be 1 over X now we can send X to infinity and these values as X gets infinitely large this should approach zero this should approach zero this is staying a nine and this is getting infinitely large and so on this problem the numerator is approaching 2 and the denominator is getting infinitely large which should make this limit approach zero all right now let's do the same problem a different way multiply top and bottom by one over x squared 2x times 1 over x squared is 2 over X 1 over x squared times one is one over x squared 1 over x squared times three x squared is three times 9x is 9 over X and times 1 is 1 over x squared now again we can send X to Infinity in this case when we do it this is going to zero this is going to zero so the whole top is heading to zero this is going to zero this is going to zero and this stays at 3 so the whole bottom is approaching three now if the top goes to zero and the bottom goes to 3 0 over 3 should be approaching zero so in either case I have an answer of zero but it was found two different ways so with rational functions a rational function by definition is one polynomial divided by another polynomial I call them n of X and D of X for the numerator and denominator polynomials there's three cases the degree where the numerator and the degree of the denominator are equal that's what we saw in our first two examples down here there is the case where the degree of the numerator is less than the degree of the denominator that's what we just saw in the last example and then there's cases where the degree of the numerator is bigger than the degree of the denominator and we'll see some of those in the upcoming videos so when the numerator is smaller than the denominator in terms of degree we have a horizontal asymptote at y equals zero when the degree of the numerator and denominator are the same we have a horizontal asymptote but it is somewhere else and it is usually found at the ratio of the coefficients so for example in this problem right up here we have 5x squared over 3x squared and we were approaching 5 3. in this next one we had 1X over 1X so we approached 1 over 1 which was one now in a calculus course you cannot just say because the degrees are the same I know the horizontal asymptote you still need to actually find it using and doing limit work the same thing with this one you can't just say the numerator is smaller than the denominator therefore the horizontal asymptote is at zero you have to actually do the limit work and verify these things