Trig limit using double angle identity | Limits and continuity | AP Calculus AB | Khan Academy

All right, let's see if we can find the limit of one over the square root of two sine of theta over cosine of two theta as theta approaches negative pi over four. Like always, try to give it a shot before we go through it together.

Well, one take on it is, well, 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 + \sqrt{2} \sin \theta) over the limit as theta approaches negative pi over four. Make sure we can see that negative there of cosine of two theta.

Both of these expressions, if these were function definitions or if we were to graph (y = 1 + \sqrt{2} \sin \theta) or (y = \cos(2 \theta)), we would get continuous functions, especially at theta equals negative pi over 4. So we could just substitute in. We say, well, this is going to be equal to this expression evaluated at negative pi over 4.

So, (1 + \sqrt{2} \sin(-\frac{\pi}{4})) over (\cos(2 \cdot -\frac{\pi}{4})). Now, (\sin(-\frac{\pi}{4})) is going to be (-\frac{\sqrt{2}}{2}). So this is (-\frac{\sqrt{2}}{2}). We're assuming this is in radians. If we're thinking degrees, this would be a negative 45 degree angle, so this is one of the trig values that it's good to know.

Now, let's see. Well, actually, just rewrite it. So this is going to be equal to (1 + \sqrt{2} \cdot (-\frac{\sqrt{2}}{2})). So this is going to be (-1). That's the numerator over here. All of this stuff simplifies to (-1) over this is going to be (\cos(-\frac{\pi}{2})).

Right? This is negative pi over 2. If you thought of it in degrees, that's going to be negative 90 degrees. Well, the cosine of that is just going to be 0. So what we end up with is equal to (\frac{0}{0}).

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 that 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 equals negative pi over 4, and it does not create this indeterminate form. We'll see other tools in our toolkit in the future.

So, let me algebraically manipulate this a little bit. If I have (1 + \frac{\sqrt{2} \sin \theta}{\cos(2 \theta)}), as you can imagine, the things that might be useful here are trig identities. 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 (\cos(2 \theta) = \cos^2(\theta) - \sin^2(\theta)), which is equal to (1 - 2 \sin^2(\theta)) and is equal to (2 \cos^2(\theta) - 1). You can go from this one to this one to this one just using the Pythagorean identity, and 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 to cancel things out that are making us get this (\frac{0}{0}).

If I could factor this into something that involves (1 + \sqrt{2} \sin \theta), I'm going to be in business. It looks like this right over here can be factored as (1 + \sqrt{2} \sin \theta \times (1 - \sqrt{2} \sin \theta)).

So, let me use this; (\cos(2 \theta)) is the same thing as (1 - 2 \sin^2 \theta), which is just a difference of squares. We can rewrite that as (a^2 - b^2) is (a + b) times (a - b). So, I can just replace this with (1 + \sqrt{2} \sin \theta) times (1 - \sqrt{2} \sin \theta).

Now, we have some nice canceling or potential canceling that can occur. So, we could say that cancels with that, and we could say that that is going to be equal. 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 are left with (1 - \sqrt{2} \sin \theta).

If we want these expressions to truly be equal, we would have to say that this one is also not defined at theta equals negative pi over 4. And actually, other places, but let’s just say (\theta \neq -\frac{\pi}{4}).

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 for this particular case, well, let's just say everything we're doing is in the open interval between, say, -1 and 1.

I think that covers it because if we have (\frac{\pi}{4}), that is not going to get us the (\frac{0}{0}) form. And (\frac{\pi}{4}) would make this denominator equal to 0. But it also makes, let's see, (\frac{\pi}{4}) will also make this denominator equal to 0 because we would get (1 - 1).

So, I think we're good if we're just assuming or restricting to this open interval. That’s okay because we're taking the limit as theta approaches something within this open interval. 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 as much trouble to put 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 as theta approaches (-\frac{\pi}{4}) of this thing without the restriction of (\frac{1}{1 - \sqrt{2} \sin \theta}). If we're dealing with this in this open interval, or actually disregarding that, this expression is continuous at it is defined and is continuous at theta equals negative pi over four.

So this is just going to be equal to (\frac{1}{1 - \sqrt{2} \cdot \sin(-\frac{\pi}{4})}). We’ve already seen that (\sin(-\frac{\pi}{4})) is (-\frac{\sqrt{2}}{2}).

So, this is going to be equal to (\frac{1}{1 - \sqrt{2} \cdot (-\frac{\sqrt{2}}{2})}). Negative, negative; you get a positive. The square root of 2 times the square root of 2 is 2, over 2 is going to be 1. So this is going to be equal to (\frac{1}{2}).

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 for dealing in this open interval except at theta equals negative pi over four. 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 that is equal to our original expression at all values of theta 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 (\frac{1}{2}), then this limit is going to be (\frac{1}{2}). 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 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. What allows you to do this 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 for all x except for a, then the limit...

Let me write it this way: equal for all x except a, and f continuous at a, then the limit of (f(x)) as (x) approaches (a) is going to be equal to the limit of (g(x)) as (x) approaches (a). I've 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 (\frac{1}{2}).