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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

• 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