we're going to look at limit laws so we're going to end up taking limits algebraically but it would be nice to have shortcuts and certain properties we can use to make our limit solving easier so for the next few slides I'm going to let C be a constant which is just any number I'm also going to let f ofx and g of X be functions where the limit as X goes to a of f ofx is equal to F and the limit as X goes to a of G ofx is equal to G and of course I'll reiterate this when we get to the important laws that require these things but these laws I introduce are going to make things much easier for us to deal with algebraically and of course you would prove these in a later analysis course but for this course we won't prove them we'll just show them with an example so the first law says if we have the limit as X goes to a of f ofx plus G ofx so two functions put together this is just the same thing as taking the limit as X goes to a of F ofx and then adding the limit as X goes to a of G ofx so we could break up a complicated function into two different functions and take the limit of them separately or we can take two limits and merge them together and take a limit together so here's an example the limit is X goes to 2 of x^2 + x + x + 1 and here I've very clearly shown them as f ofx and g ofx so on the right hand side I'm going to take the limits of them separately and on the left hand side I'm going to take the limit together and show that they're the same so on the right hand side what we're saying is that we can break it up into two limits so we can say the limit as X goes to 2 of x^2 + x and then we can add the limit as X goes to 2 of x + 1 and in all of these examples we can really simply just plug in the value for X so on the right hand side that would just be 2^ 2 + 2 and then we'd add that to the Limit as X goes to 2 of x + 1 again we just put in 2 for x and that's going to be 2 + 1 okay let's add these together so this will be 4 + 2+ 3 so our total will be 6 + 3 which is 9 so that's taking the limit separately but let's add these functions together so let's simplify this first x^2 + x + x + 1 well this is just the same thing as the limit as X goes to 2 of x^2 we group our X's together and make 2X and then add one so now we've combined the functions this is f ofx plus G of X now and let's take the limit as X goes to two so we can just plug in two so this will be 2 2 ^ 2 + 2 * 2 + 1 which is just equal to 4 + 4 + 1 which is equal to 9 so these are equal and in a later course you would prove the limit laws using the definition formally but we can see here that if we have two functions we take the same limit then we can take the limit either together or we can take them separately and it'll come out exactly the same the second law is multiplication law essentially if you have the limit as X goes to a of some number times a function so again this is just a constant it's any number then it's just equal to that constant times the limit as X goes to a of f ofx so we can just pull the constant out of the limit so here's an example the limit is X go to 2 of 3x^2 + 3 if we just put in two and evaluate it like this then we'll have three * 2^ 2 + 3 which is equal to 3 * 4 + 3 which is just 12 + 3 which is 15 but another way we can do it is we can pull out this constant factor of three so now this would be 3 * the limit as X goes to 2 of x^2 + 1 so this is equal to 3 * well let's put in two now so this will be 2^ 2 + 1 which is just 3 * well 2^ 2 + 1 is the same thing as 5 so 3 * 5 is just 15 and again we can see that these values are the same so sometimes it's easier to pull out a constant Factor because it makes multiplying all those other numbers uh much easier than just including them into every term so that's a nice multiplication law for limits the third law is that if we multiply two functions together and take their limits we can just take their limits separately and then multiply the limits together so here's an example the limit is X goes to 2 of x^2 + 2x + 1 well we can Factor this so let's Factor this this is just the limit as X goes to two of well this would be x + 1 * x + 1 okay so this kind of fits our pattern up here so F ofx is X + one and G of X is x + 1 so now we can split these up so we can just take the limit as X goes to 2 of x + 1 and then we can multiply it by the limit as X goes to 2 of x + 1 and this should be the same thing so the limit is X goes to 2 of x + 1 1 is just going to be 2 + 1 we'll multiply that by the limit as X goes to 2 of x + 1 which is also just 2 + 1 so this will end up being 3 * 3 which is 9 and if we just plug two in into all of our X values here which we've already done before this would be 4 + 4 + 1 which is just equal to 9 so we can see that even if we multiply two functions together and take their limits we can just take the limit separately and then multiply them together later which if you're dealing with something like X 5th x to the 6th it might be easier to break them down into two factors of X cubed or maybe an X squ and an X cubed and then take the limit separately just to make it easier for your computation so this is a limit law you can use to make your computations a little bit easier on an exam if you see that it can be done the last law I'll talk about is the division law so the limit as X goes to a a of f ofx over G ofx is just the same thing as the limit as X goes to a of f ofx over the limit as X goes to a of G ofx but that's only if the limit as X goes to a of G of X is not equal to zero in other words what this says is don't divide by zero otherwise it's bad so here's the example the limit as X goes to 2 of 3x^2 + 1 if I just plug in two for these values this will be 3 * 2 over 2^ 2 + 1 which is just equal to 6 over 5 so let's just keep this in mind that if we evaluate it like this this is just equal to 6 over 5 but we can take the limit individually so we can take the limit as X goes to 2 of 3x and we can divide that by the limit as X goes to 2 of x 2 + 1 and if we plug in some values we see 3 * 2 is just 6 and if we plug two in for X on the bottom it'll be 2^2 + 1 which is equal to 5 and these are exactly the same so again these might be really obvious to you but there's a lot of mathematics behind it that we have to formally Define in order to prove these things we're just taking them H for granted for now uh but these are really important rules that you can use when you have to do your computation questions and mechanical questions on your exams in your course so here's a practice question and I'm going to do all these laws individually just so we can see how they work so I can take everything at once I can just plug five into these values except we're going to get some big numbers to deal with I mean they're going to be big regardless but uh let's just show how we can use limit laws to simplify this a little bit so the first thing I can do is I can use the mul multiplication law and I can pull out a factor of five from everything so I can just take the five out and then this would be the limit as X goes to 5 of well this 10 x^2 would become 2x^2 the 5x would just become X and the minus 20 would become Min -4 okay the next thing I can do is I can take all of these limits individually instead so I can do it five * and then this will be the limit as X approaches 5 of 2x^ sared plus the limit as X approaches 5 of X plus the limit as X approaches five of 4 okay so at this point you might be thinking maybe we should just do it the original way it's faster and indeed it is faster but just to really demonst demonstrate these laws I'm going to do it step by step like this so this will be five times Well in the first case we can pull out a two so this would be 2 * the limit as X goes to 5 of x^2 plus well the limit as X goes to 5 of X is just five what about the limit as X goes to 5 of four well there's no X there to plug into so it's just going to be four okay so this will be five * well so we're going to have two times the limit as X goes to 5 of x^2 what is the limit as X goes to 5 of x^2 that'll be 5^ 2 so this will be 2 * 25 because 5^ s is 25 then + 5 + 4 that's just + 9 so this will now be 5 * well 2 * 25 is 50 + 9 is 59 and now we can see this will be 5 time 59 which is just 250 + 45 so this should be 295 and that's showing all the limit laws of course it really is faster just to plug in 5 and X up top so this would be 10 * 5^ 2ar + 5 * 5 minus 20 uh okay that is make a mistake here so this should be Min -4 in all of these this should be + one which should be 51 which should be 255 okay I did of course I missed a negative sign so this would be 10 * 25 which is 250+ 5 * 5 which is 25 - 20 so this would be 275 - 20 which is just 255 so we can see that they're both exactly the same one is quicker of course but the other is taking each limit law and doing it step by step which of course with more complicated questions will be very useful if you have any questions please leave them in the comments below and I will do my best to answer them