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Current time:0:00Total duration:4:05

Limit of (1-cos(x))/x as x approaches 0

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.E (LO)
,
LIM‑1.E.2 (EK)

Video transcript

what we want to do in this video is figure out what the limit as X approaches 0 of 1 minus cosine of X over X is equal to and we're going to assume we know one thing ahead of time we're going to assume we know that the limit as X approaches 0 of sine of X over X that this is equal to 1 and I'm not going to reprove this in this video but we have a whole other video dedicated to proving this a famous limit and we do it using the squeeze or the sandwich theorem so let's see if we can work this out so the first thing we're going to do is algebraically manipulate this expression a little bit what I'm going to do is I'm going to multiply both the numerator and the denominator by 1 plus cosine of X so times the Dominator I have to do the same thing 1 plus cosine of X I'm not changing the value of the expression this is just multiplying it by 1 but what does that do for us well I can rewrite the whole thing as the limit as X approaches 0 so 1 minus cosine of X times 1 plus cosine of X well that is just going to be let me do this in another color that is going to be 1 squared which is just 1 minus cosine squared of X cosine squared of X difference of squares and then in the denominator I am going to have these which is just x times 1 plus cosine of X now what is 1 minus cosine squared of X well this comes straight out of the Pythagorean identity trig identity this is the same thing as sine squared of X so sine squared of X and so I can rewrite all of this as being equal to the limit as X approaches 0 and let me rewrite this as instead of sine squared of X if that's the same thing as sine of X times sine of X I'll write that way sine x times sine X so I'll take the first sine of X so I'll do I'll take this one right over here and put it over this X so sine of X over X times the second sine of X let's say this 1 over 1 plus cosine of X times sine of X over 1 plus cosine of X all I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation well here the limit of the product of these two expressions is going to be the same thing as the product of the limits so I can rewrite this as being equal to the limit as X approaches 0 of sine of X over x times the limit as X approaches 0 of sine of X over 1 plus cosine of X now we said going into this video that we're going to assume what we know that we know what this is we prove it in other videos what is the limit as X approaches 0 of sine of X over X well that is equal to 1 so this whole limit is just going to be dependent on whatever this is equal to well this is pretty straightforward here as X approaches 0 the numerator is approaching 0 sine of 0 is 0 the denominator is approaching cosine of 0 is 1 so the denominator is approaching 2 so this is approaching 0 over 2 or just 0 so that's approaching 0 1 times 0 well this is just going to be equal to 0 and we're done using that fact and a little bit of trig identities and a little bit of algebraic manipulation we were able to show that our original limit the limit as X approaches 0 of 1 minus cosine of X over X is equal to 0 and encourage you to graph it you will see that that makes sense from a graphical point of view as well