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## Limits using algebraic manipulation

Current time:0:00Total duration:10:31

# Trig limit using double angle identity

AP.CALC:

LIM‑1 (EU)

, LIM‑1.E (LO)

, LIM‑1.E.1 (EK)

## Video transcript

- [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 or notebook, you wouldn't be taking, putting, or taking as much trouble to be putting all of these caveats in. So, what we've now realized is that, okay, this expression, actually, let's think about this. Let's think about the limit, the limit as theta
approaches negative pi over 4 of this thing, without the restriction, of 1 over 1 minus the square root of 2 sine of theta. If we're dealing with this over, you know, with this open interval, wait, actually, even disregarding that, theta, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over 4 so this is just going to be equal to 1 over 1 minus the square root of 2 times sine of negative pi over 4. Sine of negative pi over 4. Sine of negative pi over 4, we've already seen is negative
square root of 2 over 2, and so this is going to be equal to 1 over 1 minus square root of 2 times the negative square root of 2 over 2, so negative, negative, you get a positive, square root of 2 times square root of 2 is 2, over 2 is going to be 1. So this is going to be equal to 1/2. And so, I want to be very clear. This expression is not the same thing as this expression. They are the same thing
at all values of theta, especially if we're dealing
in this open interval except at theta equals negative pi over 4. This one is not defined and this one is defined, but as we've seen multiple times before, if we find a function that is equal to our original or an expression, is equal to our original expression, and all values of theta except, except where the original
one was not defined at a certain point, but this new one is defined
and is continuous there, well then these two limits
are going to be equal, so if this limit is 1/2, then this limit is going to be 1/2, and I've said this in previous videos. It might be very tempting to say, well, I'm just going to algebraically
simplify this in some way to get this, and I'm not going to
worry about too much about these constraints, and then
I'm just going to substitute negative pi over 4, and you will get this answer which is the correct answer but it's really important to recognize that this expression and this expression are not the same thing and what allows you to do this is, is the truth that if you have two functions, if you have f and g, two functions equal, let me write it this way, equal, equal for all x, except for all, wait, let me just write this this way, for all x except for a, then the limit, then, and let me write it this way, equal for all except, for all x except a and f continuous, continuous at a, then, then the limit of f of x as x approaches a is going to be equal
to the limit of g of x as x approaches a, and I said this in multiple videos and that's what we are doing right here, but just so you can make
sure you got it right, the answer here is 1/2.