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  • - [Voiceover] All right, let's see if we can find the limit

  • of 1 over the square root of 2 sine of theta

  • over cosine of 2 theta,

  • as theta approaches negative pi over 4.

  • And like always, try to give it a shot before we go through

  • it together.

  • Well, one take on it is well, let's just,

  • let's just say that this is going to be the same thing

  • as the limit,

  • as theta approaches negative pi over 4

  • of 1 plus square root of 2 sine theta

  • over the limit

  • as theta approaches negative pi over 4.

  • Make sure we can see that negative there,

  • of cosine of 2 theta,

  • and both of these expressions are,

  • if these were function definitions

  • or if we were to graph y equals

  • 1 plus square root of sine,

  • square root of 2 times sine theta,

  • or y equals cosine of 2 theta,

  • we would get continuous functions,

  • especially at theta is equal to negative pi over 4,

  • so we could just substitute in.

  • We'll see well this is going to be equal to

  • this expression evaluated at negative pi over 4,

  • so 1 plus square root of 2

  • times sine of negative pi over 4,

  • over cosine

  • of 2 times negative pi over 4.

  • Now, negative pi over 4,

  • sine of negative pi over 4 is going to be

  • negative square root of 2 over 2,

  • so this is negative square root of 2 over 2,

  • we're assuming this is in radians,

  • if we're thinking in degrees, this would be

  • a negative 45-degree angle, so this is one of the,

  • one of the trig values that it's good to know

  • and so if you have,

  • if you have 1,

  • so let's see,

  • actually, let me just rewrite it,

  • so this is going to be equal to 1 plus

  • square root of 2 times that

  • is going to be negative 2 over 2,

  • so this is going to be

  • minus 1,

  • that's the numerator over here.

  • All of this stuff simplifies to negative 1

  • over,

  • this is going to be cosine of

  • negative pi over 2, right?

  • This is negative pi over 2,

  • cosine of negative pi over 2, if you thought in degrees,

  • that's going to be negative 90 degrees.

  • Well, cosine of that is just going to be zero,

  • so what we end up with

  • is equal to zero over zero,

  • and as we've talked about before,

  • if we had something non-zero divided by zero,

  • we'd say, okay, that's undefined.

  • We might as well give up,

  • but we have this indeterminate form, it does not mean

  • the limit does not exist.

  • It's usually a clue that we should use some tools

  • in our toolkit, one of which is to do some manipulation

  • here to get an expression that maybe is defined

  • at theta is equal to,

  • or does not,

  • is not an indeterminate form,

  • that theta is equal to pi over 4

  • and we'll see other tools in our toolkit in the future.

  • So let me algebraically manipulate this a little bit.

  • So if I have 1 plus the square root of 2,

  • sine theta, over cosine 2 theta,

  • as you can imagine,

  • the things that might be useful here are our trig identities

  • and in particular, cosine of 2 theta seems interesting.

  • Let me write some trig identities involving cosine

  • of 2 theta.

  • I'll write it over here.

  • So we know that cosine

  • of 2 theta

  • is equal to

  • cosine squared of theta

  • minus sine squared of theta

  • which is equal to 1

  • minus 2 sine squared of theta

  • which is equal to

  • 2 cosine squared theta minus 1,

  • and you can go from this one to this one to this one

  • just using the Pythagorean identity.

  • We proved that in earlier videos in trigonometry

  • on Khan Academy.

  • Now, do any of these look useful?

  • Well, all of these three are going to be differences

  • of squares, so we can factor them in interesting ways,

  • and remember, our goal at the end of the day

  • is maybe cancel things out that are making us get this

  • zero over zero,

  • and if I could factor this

  • into something that involved a 1 plus square root

  • of 2 sine theta,

  • then I'm going to be in business,

  • and it looks like,

  • it looks like this right over here,

  • that can be factored as

  • 1 plus square root of 2 sine theta

  • times

  • 1 minus square root of 2 sine theta,

  • so let me use this.

  • Cosine of 2 theta is the same thing,

  • cosine of 2 theta is the same thing

  • as 1 minus 2 sine squared theta,

  • which is just a difference of squares.

  • We can rewrite that as,

  • this is a-squared minus b-squared,

  • this is a plus b times a minus b,

  • so I can just replace this with

  • 1 plus square root of 2 sine theta

  • times 1 minus square root of 2 sine theta,

  • and now, we have some nice cancelling,

  • or potential cancelling that can occur,

  • so we could say

  • that cancels with that

  • and we could say that that is going to be equal,

  • and let me do this in a new color,

  • this is going to be equal to,

  • in the numerator we just have 1,

  • in the denominator we just are left with

  • 1 minus square root of 2 sine theta,

  • and if we want these expressions to truly be equal,

  • we would have to have them to have the same,

  • if you view them as function definitions,

  • as having the same domain,

  • so this one right over here,

  • this one we already saw is

  • not defined at theta is equal to negative pi over 4,

  • and so this one,

  • in order for these to be equivalent,

  • we have to say that this one is also not,

  • and actually, other places, but let's just,

  • let's just say theta

  • does not,

  • does not equal negative,

  • negative pi over 4,

  • and we could think about all of this happening

  • in some type of an open interval around negative pi over 4

  • if we wanted to get very precise,

  • but if we wanted to,

  • for this particular case,

  • well, let's just say,

  • everything we're doing is in the open interval,

  • so in,

  • in open interval,

  • in open interval between

  • theta,

  • or, say, negative 1 and 1,

  • and I think that covers it

  • because if we have pi,

  • if we have pi over 4

  • that is not going to get us the

  • zero over zero form,

  • and pi over 4

  • would make this denominator equal to zero

  • but it also makes,

  • let's see, pi over 4 also will make this denominator

  • equal to zero,

  • 'cause we would get 1 minus 1,

  • so I think,

  • I think we're good if we're just assuming,

  • if we're restricted to this open interval

  • and that's okay because we're taking the limit

  • as it approaches something within this open interval,

  • and I'm being extra precise because I'm trying to explain it

  • to you and it's important to be precise,

  • but obviously, if you're working this out on a test