episode four techniques for computing limits so spend today talking about different ways that we can find the values of limits so we'll start with some that are really easy some of them should be just incredibly easy if you just read them aloud all right so let's assume C is a constant well the limit as X approaches a of C well that's gonna be C right C is a constant it does not care what Dex is doing and so the limit is just C another one that should be pretty easy as a limit as X approaches a of X what does X approach as X approaches a it approaches a right so this limit would be a couple of others that are easy maybe not as easy as those last two but they they make sense well let's say that the limit as X approaches a of f of X is equal to L and the limit as X approaches a of G of X is equal to K and C is a constant so we'll start with the limit as X approaches a of C of f of X right well that would just be C times whatever the limit as X approaches a of f of X is its though BC times L we can bring a constant multiple outside of the limit operator I think you would probably Intuit this but it's a good idea to think about this explicitly and then similarly if we take the limit as X approaches a of f of X plus or minus G of X we can just do those limits separately provided they exist then though that limit would just be the limit of f of X plus or minus the limit of G of X so the L plus or minus K I hope this feels obvious to you these two properties are often paired together when we're talking about an operator like the idea of taking the limit as X approaches a it's a linear operator something that satisfies these two properties is a linear operator and we'll see that again throughout the semester there's some similar other ways to combine some limits we need to take the limit as X approaches a f of X times G of X well it's what we think it is right if we know the limit of f and we know the limit of G then the limit of their product is just the product of their limits and same thing for division provided we don't divide by zero so the limit as X approaches I have f of X divided by G of X would be the limit of the top divided by the limit at the bottom unless the limit at the bottom is zero so BL over K and n is a positive integer we have a function raised to the N power well we can just move that limit operator inside the power and so it's going to be the limit of f of X as X approaches a that's L to the nth power and then the same thing for the enth root as long as we can move the limit operator inside the root and as long as the limit is positive when n is even we can take the enth root of 0 but we do run into a little bit of a problem with the limit trying to do that in the context of a limit the limits of compositions of functions work in a similar way right really that last one was just a subset of this idea if the limit as X approaches a of G of X is L and the limit as X approaches L of f of X is just F of L and then the limit as X approaches a of f of G of X is right these are gonna be pretty obvious here if we just plug things in this is what we'll get and the limit of trig functions are actually pretty nice too as long as it a is in the domain of the given function then the limit as X approaches a of that trig functions just that trig function evaluated today right so the limit as X approaches a and the sine of X is the sine of a same thing for cotangent and cosecant all of those we got to remember what their domains are that's a separate issue but as long as we can remember what their domains are evaluating limits in that domain those are easy polynomials you're also very easy now right we looked at this before the limit is X approaches to of 3x plus one we plugged in some values and saw that the limit was probably seven and we now have the ability to show that it actually in seven right really the best way to think about this is hey I have the sum of two things three times X plus one so I would think of that as the limit as X approaches 2 of 3x plus the limit as X approaches 2 of one and well those are pretty easy right we just plug 2 into that and we get 7 the limit as X approaches 3 of 2x cubed minus x squared plus 6x minus 2 we can split up the additions and we can factor out constants and use powers and we get that this is 2 times 3 cubed minus 3 squared plus 6 times 3 minus 2 that's 61 in general if P of X is a polynomial evaluating limit is really easy we just plug in that value the limit as X approaches a of P of X is equal to P of a for any polynomial rational functions are basically just as easy right a rational function is a polynomial divided by another polynomial and so as long as that polynomial on the bottom is not 0 when we plug it in we have our limit so the limit as X approaches 1 of x squared minus 4 divided by x squared plus 3x plus 7 that's a rational function so I just plug in 1 and we get negative 3 on the top and we get 11 on the bottom 11 is not zero so this is minus 311 in general just P of x over Q of X is a rational function the limit as X approaches a of P of X divided by Q of X is equal to P of a divided by Q of a as long as Q of a is not 0 QA where 0 and P of a was nonzero we'd have an infinite limit and a vertical asymptote if both P of a and Q of a or zero that's another topic we'll talk about that in a second so some limits pretty easy as long as we can plug in and get a number that's pointing to be the limit right so the limit as X approaches 5 of X plus 2 square root of x divided by x squared minus 10x plus 1 try to plug in 5 in the top we get 5 plus 2 square root of 5 in the bottom we get 25 minus 50 plus 1 so that's minus 5 plus 2 square root of 5 divided by 24 that's the limit no problems there what about the limit is Theta approaches PI over 6 of 1 plus the sine of theta divided by 2 plus the secant of theta well PI over 6 is in the domain of these two trig functions so as long as we don't end up with 0 in the bottom we don't have any troubles here all right so this is just one plus the sine of PI over 6 divided by 2 plus the secant of PI over 6 we got to remember our trig values all right that one's gonna be 1 plus 1/2 divided by 2 plus 2 over the square root of 3 both of those come from the unit circle and secant is 1 over cosine we can simplify that down we get it 3 ABS / thing in the bottom of the common denominator flip and multiply I would be perfectly fine with the first expression there 3 times the square root of 3 / 4 times 1 plus the square root of 3 I'm perfectly fine with that I don't need you to rationalize the denominator but I need you to know how to do it so expecting to know how to show that this is actually equal to 3 times 3 minus the square root of 3 divided by 8 you just multiply by the conjugate we'll talk a little bit more about this next time but you don't need to actually do it most of the time how about this one the limit as X approaches negative 1 of 7 divided by X to the 4th plus 5x cubed minus x squared plus 3x plus 2 let me just try to plug in negative 1 as long as we don't get 0 on the bottom we don't have any trouble all right that's gonna be 1 minus 5 minus 1 minus 3 plus 2 that ends up with negative 6 on the bottom so this is negative 7/6 no problems there one word 2 is some trig values the limit as X approaches PI of 2 plus X cosine X divided by 8 plus sine squared X again you just plug in PI all right so we get 2 plus PI times the cosine of pi divided by 8 plus the sine squared of pi now remember what cosine of PI is Cristiano PI is negative 1 and the sine of pi is 0 so this is 2 minus pi divided by 8 so let's talk about some cases that aren't covered in what we just looked at and come up with a way to figure out something really around that but the technique that we're going to use relies on this idea of functions that disagree only at one point let f of X equal G of X when x is not equal to a in an open interval containing a what happens at a doesn't matter it just matters that on either side of a there's a little bit of room where the two functions are equal to each other and say that we know the limit as X approaches a of G of X it exists or if it were infinity or negative infinity if we know what the limit as X approaches a of G of X is then we also know what the limit as X approaches a of f of X is if they agree everywhere except for a their limits have to be the same as X approaches a so where this is very helpful isn't a 0/0 situation 0/0 we can't conclude anything we have to do some more work and this is the function that that helps us get around that where this comes in handy most often is with rational functions we would factor and cancel it a couple of other options we'll talk about next time all right so let's look at the limit as X approaches 2 of 4x squared minus 11x plus 6 divided by X minus 2 first thing I try to do is plug in plug in 2 and I get 0/0 so I should try to factor it bottom is pretty easy to factor right when X is not to the bottom it just is what it is now in the top I plugged two in and got zero the fundamental theorem of algebra says that X minus 2 must be a factor I need to figure out what the other one is right X minus 2 is one factor so four X minus three needs to be the other factor you can double check by foiling that out right 4x times X would be 4x squared 4x times minus 2 would be minus 8x and then we have minus 3x that's minus 11x minus 3 times minus 2 is plus 6 right so hey what do we have here hey that X minus 2 divided by X minus 2 that's my 0 over 0 problem right long as X is not 2 I can cancel those and so when X is not 2 the function that I was trying to take the limit of is equal to 4x minus 3 but I don't care what happens when X is 2 right I'm taking the limit as X approaches 2 so these two functions 4x squared minus 11x plus 6 divided by X minus 2 that function is equal to the simpler function 4x minus 3 as long as X is not 2 so I can take the limit as X approaches 2 of 4x minus 3 that's easy to do I just plug in 2 and I get 5 and so the limit that I was interested in is equal to that limit so the limit I was interested in is equal to 5 let's try another one the limit as X approaches 1 of 3 x squared minus 4x plus 1 divided by x squared plus 4x minus 5 so we just plug in x equals 1 and in the top we get 3 minus 4 plus 1 that's 0 and the bottom we get 1 plus 4 minus 5 also 0 so 0 divided by 0 I should try to factor it because I plugged in 1 and got 0 at both the top and the bottom I know X minus 1 will be a factor in both the top and the bottom that it has to be that way for me to cancel the zero over zero problem so now I just need to figure out what the other two factors aren't right so well on the top it's going to start with a 3x so I get 3x times X equals 3x squared and then the other one will be negative one you can double-check that in the bottom I know that the it needs to start with X so that I get x squared there and then I need a plus five to multiply by negative one and get minus five double check the outside and inside terms and you get plus 4x hey now I can cancel the zero over zero problem right so the limit that I was interested in is equal to the limit as X approaches 1 of 3x minus 1 divided by X plus 5 and that one I can do just by plugging in all right it's gonna be two divided by six so that's one third so this limit is one third when you plug in and get a zero over zero limit try factoring and canceling if it's a rational function it will work we will do some more examples of this next time