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Current time:0:00Total duration:7:02

AP.CALC:

LIM‑1 (EU)

, LIM‑1.E (LO)

, LIM‑1.E.1 (EK)

let's see if we can find the limit as theta approaches 0 of 1 minus cosine theta over 2 sine squared theta and like always pause the video and see if you could work through this alright well our first temptation is to say well this is going to be the same thing as the limit of 1 minus cosine theta as X approaches or not X as theta approaches zero theta as theta approaches zero over the limit as theta approaches zero of 2 sine squared theta now both of these expressions which could be used to define a function that they would be continuous if you graph them they'd be continuous at theta equals 0 so the limit is going to be the same thing as just evaluating them at theta equals 0 so this is going to be equal to 1 minus cosine of 0 over 2 sine squared of 0 now cosine of 0 is 1 and then 1 minus 1 is 0 and sine of 0 is 0 and you square it you still got 0 and you're multiplied times - you still got 0 so you got 0 over 0 so once again we have that indeterminate form and once again this indeterminate form when you have 0 over 0 doesn't mean to give up it doesn't mean that the limit doesn't exist it just means well maybe there's some other approaches here to work on if you've got some nonzero number divided by 0 then you say okay that limit doesn't exist and you'd say well you just say it doesn't exist but let's see what we can do to maybe to maybe think about this expression in a different way so if we said so let's just say that this let me do some other colors here let's say that this right over here is f of X so f of X is equal to 1 minus cosine theta over 2 sine squared theta and let's see if we can rewrite it in some way that at least the limit as theta approaches 0 isn't going that we're not going to get the same 0 over 0 well we can we got some trig functions here so maybe we can use some of our trig identities to simplify this and the one that jumps out at me is that we have the sine squared of theta and we know from the Pythagorean Pythagorean identity in trigonometry comes straight out of the unit circle definition of sine cosine we know that we know that sine squared theta plus cosine squared theta is equal to 1 or we know that sine squared theta is 1 minus cosine squared theta one minus cosine squared theta so we could rewrite this this is equal to 1 minus cosine theta over 2 times 1 minus cosine squared theta now this is a 1 minus cosine theta this is a 1 minus cosine squared theta so it's not completely obviously out of how you can simplify it until you realize that this could be viewed as a difference of squares if you view this as if you view this as a squared minus B squared we know that this can be factored as a plus B times a minus B so I could rewrite this this is equal to 1 minus cosine theta over 2 times I could write this as 1 plus cosine theta times 1 minus cosine theta 1 plus cosine theta times 1 minus 1 minus cosine theta and now this is interesting I have a 1 minus cosine theta in the numerator and I have a 1 minus cosine theta in the denominator now we might be tempted to say well let's just let's just cross that out with that and we would get we would simplify it and get f of X is equal to 1 over and we could distribute this 2 now we could say 2 plus 2 cosine theta we could say well they aren't D is the same thing and we would be almost right because f of X this one right over here this this is defined this this right over here is defined when theta is equal to 0 well this one is not defined when theta is equal to 0 when theta is equal to 0 you have a zero in the denominator and so what we need to do in order for this f of X or in order to for this to be the same thing we have to say theta cannot be equal to zero but now let's think about the limit again essentially what we want to do is we want to find the limit as theta approaches zero of f of X and we can't just do direct substitution into if we really take this seriously because we're going to like oh well if I try to put zero here it says theta cannot be equal to zero f of X is not defined at zero this expression is defined at zero but this tells me well I really shouldn't apply zero to this function but we know that if we can find another function that is defined that is the exact same thing as f of X except at zero and it is continuous at zero and so we could say G of X is equal to one over two plus two cosine theta well then we know this limit is going to be the exact same thing as the limit of G of X as theta approaches zero once again these two functions are identical except f of X is not defined at theta equals the zero while G of X is but the limits as theta approaches zero are going to be the same and we've seen that in previous videos and I know what a lot of you thinking how this seems like a very you know why don't I just you know do this algebra here cross these things out and there's a substitute 0 for theta well you could do that and you would get the answer but you need to be clear or it's important to be mathematically clear of what you are doing if you do that if you if you just cross these two out and all of a sudden your expression becomes defined at 0 you are you are now dealing with a different expression or a different function definition so to be clear if you want to say this is the function you're finding the limit of you have to put this constraint in to make sure it has the exact same domain but lucky for us we can say if we found another another function that's continuous at that point that doesn't have that gap there that doesn't have that point discontinuity out the limits are going to be equivalent so the limit is theta approaches 0 of G of X well that's just going to be since it's continuous at 0 we could say that's just going to be we can just substitute that's going to be equal to G of zero which is equal to one over two plus cosine to one over two plus two times cosine of zero cosine of zero is one so it's just 1 over 2 plus two which is equal to you a little bit of a drumroll here which is equal to 1/4 and we are done

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