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QUESTION 1 [6]Evaluate lim 1-0if it exists_QUESTION 2 [6] Find an expression for cos 40 and sin 40 in terms of sin 0 and cos 0,...

Question

QUESTION 1 [6]Evaluate lim 1-0if it exists_QUESTION 2 [6] Find an expression for cos 40 and sin 40 in terms of sin 0 and cos 0,

QUESTION 1 [6] Evaluate lim 1-0 if it exists_ QUESTION 2 [6] Find an expression for cos 40 and sin 40 in terms of sin 0 and cos 0,



Answers

Use Theorem 3. 10 to evaluate the following limits. $$\lim _{\theta \rightarrow 0} \frac{\cos ^{2} \theta-1}{\theta}$$

Limits. Eggs approaches. Zero off. Signed a four X over sign of six X right here. Right. Uh, we're just gonna make it. You just have to do a little bit of algebraic manipulations. I signed up for eggs. I'm gonna most implied by a four x for X. I'm gonna do the same here. Well supplied by six X over six X. Right. So what I'm gonna get is limits Execution. Zero of sign for X times four x four x and then over. Signed six x time six x over six X. Right. So these excess air going to cancel as the first thing. Right? So in those excess cancel that I'm gonna have is limit four signed for X over four X over the limit. You can spend the limit, right. It is just part of the, uh, properties of limits. Have just made it. Six. Signed six x over 66 Remembered at last. You told me before this one. He said that if you have something like this, where whatever is in front here is that same us. What is that about him? It's been proven by increased serum that is actually one right. So that is actually one. So if this is one, you just have one times four, which is for right. So this is before the same vein. You have this here also one one time 66 year one time six is also six. So you have this one. So then it's just about every two or three.

Yes, they were given the limits as they're approaching zero up one of my next call sign dinner over. Sign two. They don't. If your car is entity for signed to data to write that, that's just to sign up there. No close side if they don't. And then we have one minute school sign. We could be right that as two sine squared, you know, open to that's our half angle identity. Okay, now it's multiplied by date over four. Did we get the limit? Data about your zone one over to co sign it. Uh, times to sine squared data over to over. Data squared over four. Have they Don't hear sign, You know. You know, over four. No, nothing. Oh, yeah, there before. And this just simplifies to data over for plugging in. Argueta is equal to zero We get. This is equal to go over point, which is equal to go

All right, So in this video, we're asked to determine the limit as data close to zero off one minus close. I data divided by sine two theta. And will you need to use this? These relationships were here. These limits. So the limited scale goes to zero. Sign came divided by K equals one. And then what? What what we get. What immediately applies from that is that the saw and the limited scale goes to zero off K divided by sidekick equals one. So basically, the limit of the reciprocal also equals one. No first thing we're going to do, we're going to try to manipulate Well, we have inside the limit, so But we can do is we can split this into two fractions. So one times sign of truth, Ada multiplied by one minus consigned data divided by one. Now remember, we have ah to theater inside the sign So we wanted to fade in the numerator But we have a one. So what we do is we multiply by two theater divided by two theater Kohstuh theater divided by two theaters Just controverted one. So that's valid. You can you can You can vote by by one that doesn't changing. And then no one extra thing you can do so no, we Now, here we go to the next step. We have a toothache and the numerator. But another interesting thing is we can take the state of it's right here and then just with all the way to be underneath the one minus closer. So now what we have We have a 1/2 on the outside. We have a two theater on top of that sign do data and then we have one minus co sign data divided by all right. Now, that limit as this ghost is ears of the 1/2 is a course that it doesn't change. Just stays. 1/2 a state of was zero. This becomes a one and I already know from earlier one minus close. I did a divided by beta. This is zero. So we have 1/2 times one times zero and the Limited give zero

For this problem, if we plug the zero in, you can see right about you have a zero. Do you know I'm in theater which is undefined. Uh, There are no algebraic ways of doing this problem. So you can go right to locate halls role, which means you're going to do the derivative of the numerator, derivative of cosine is negative sine of X. The derivative of minus one is nothing. And then the derivative of the denominator derivative of X is one. And now we can plug in the limit value. So negative sine of 0/1. So sign of zero. You could do the sign graph And sign of zero.


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