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Current time:0:00Total duration:7:43

AP.CALC:

LIM‑4 (EU)

, LIM‑4.A (LO)

, LIM‑4.A.1 (EK)

, LIM‑4.A.2 (EK)

let's say we need to evaluate the limit the limit as X approaches zero of 2 sine of X 2 sine of X minus sine of 2 X all of that over all of that over X minus sine of X now the first thing that I always try to do when I first see a limit problem is oh hey what happens if I just try to evaluate if I just try to evaluate this function at X is equal to zero maybe nothing crazy happens so let's just try it out if we try to do x equals zero what happens we get two sine of zero which is zero minus sine of two times zero well sine of zero that's going to be sine of zero again which is zero so our numerator is going to be equal to 0 right sine of zero that's zero and then we have another sine of zero there it's another zero so all zeros and our denominator we're going to have a zero minus sine of zero well that's also going to be zero but we have that indeterminate form we have that undefined zero over zero that we talked about in the last video so maybe maybe we can use l'hopital's rule here in order to use l'hopital's rule then the limit as X approaches 0 of the derivative of this function over the derivative of this function needs to exist so let's just apply l'hopital's rule and see let's just take the derivative of each of these and see if we can find the limit if we can and that's going to be the limit of this thing so this thing assuming that it exists is going to be equal to the limit as X approaches 0 of the derivative of this numerator up here and so what's the derivative of the numerator going to be able to it in a new color I'll do it in green well the derivative of 2 sine of X is 2 cosine of X 2 cosine of X and then minus well the derivative of sine of 2 X is 2 cosine of 2 X so minus 2 cosine of 2 X just use the chain rule their derivative of the inside is just 2 that's the 2 out there derivative of the outside is sine of 2x and we had that negative number out there so that's the derivative of our numerator and what is the derivative of our denominator well derivative of X is just 1 and derivative of sine of X is just cosine of X so 1 minus cosine of X so let's try to evaluate this limit what do we get if we have put a 0 up here we're going to get 2 times cosine of 0 which is 2 let me write it like this so this is 2 times cosine of 0 which is 1 so it's 2 minus 2 cosine of 2 times 0 cosine of let me write it this way actually let me just do it this way if we just straight-up evaluate the limit of the numerator denominator what are we going to get we get 2 cosine of 0 which is 2 minus 2 times cosine of well this 2 times you are still going to be 0 so minus 2 times cosine of 0 which is 2 all of that over all of that over 1 minus the cosine of 0 which is 1 so once again we get 0 over 0 once again we get 0 over 0 so does this mean that the limit doesn't exist no it still might exist we might just want to do laa petals rule again let me take the derivative of that and put it over the derivative of that and then take the limit and maybe law petals rule will help us on the next stage so let's see what if it if it gets us anywhere so this should be equal to the limit if the lift flop oh tiles rule applies here we're not 100% sure yet this should be equal to the limit as X approaches 0 of the derivative of that thing over the derivative of that thing so what's the derivative of 2 cosine of X well derivative of 2 code well could riff cosine of X is negative sine of X so it's negative 2 sine of X and the derivative of cosine of 2 X is negative 2 sine of 2 X so we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times the 2 so it's going to be plus 4 sine of 2 X let me make sure I did that right we have the minus 2 or the negative 2 on the outside cosine of 2x is going to be two times negative sine of X so the two times two is four the negative sine of x times the negative right there is a plus there's a positive sign so it's sine of 2x so that's the numerator when you take the derivative and the denominator this is just an exercise in taking derivatives what's derivative the denominator derivative of 1 is zero and really the derivative of negative cosine of X is just well that's just sine of X sine of X so let's take this limit so this is going to be equal to well immediately if I take X is equal to 0 in the denominator I know that sine of zero is just zero let's see what happens in the numerator negative two times sine of zero that's going to be zero and then plus four times sine of two times zero well that's still sine of zero so that's still going to be zero so once again we got indeterminate form again are we done and we give up do we say that law petals rule didn't work no because this could have been our first limit problem and if this is our first limit problem say hey maybe we could use l'hopital's rule here because we got an indeterminate form and where both the numerator and the denominator approach zero as X approaches zero so let's take the derivatives again so let's take the driven this will be equal to if the limit exists the limit as X approaches zero let's take the derivative of the numerator the derivative of negative 2 sine of X is negative 2 cosine of X and then plus the derivative for sine of 2x well it's 2 times 4 which is 8 times cosine of 2x derivative of sine of 2x is 2 cosine of 2x and that 2 got that first 2 gets multiplied by the 4 to get the 8 and then the derivative of the denominator derivative of the denominator derivative of sine of X is just cosine of X so let's evaluate this character so we've made it looks like we've made some headway at least or maybe l'hopital's rule will stop applying here because we take the limit as X approaches 0 of cosine of X that is 1 so we're not we're definitely not going to get that indeterminate form that 0 over 0 on on this iteration let's see what happens to the numerator we get negative 2 times cosine zero well that's just negative - because cosine of zero is one plus eight times cosine of 2x well X is zero so it's going to be cosine of zero which is 1 so it's just going to be an 8 so negative 2 plus 8 well this thing right here negative 2 plus 8 is 6 6 over 1 this whole thing is equal to 6 so lapa tiles rule it applies to this last step if we if this was the problem we were given we said hey when we try to apply the limit we get the limit as this numerator approaches zero is zero limit as this denominator approaches zero is zero but the limit as the numerator as the derivative of the numerator over the derivative of the denominator that exists and it equals six so this limit must be equal to 6 well if this limit is equal to 6 by the same argument this limit is also going to be equal to 6 and by the same argument this limit has got to also be equal to 6 and we're done

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