Video: Applying the Squeeze Theorem

Consider the following arc of a unit circle, where ray 𝑂𝑃 is inclined at πœƒ radians. 1) What, in terms of πœƒ, are the coordinates of 𝑃? 2) Write the following inequalities in terms of sin πœƒ, πœƒ, and cos πœƒ: 𝑄𝑅< length of the arc from 𝑇 to 𝑄 < 𝑇𝑃. 3) By dividing your inequalities by sin πœƒ, using the squeeze theorem and the fact that lim_(πœƒ β†’ 0) cos πœƒ = 1, which of the following conclusions can you draw? [A] lim_(πœƒ β†’ 0) πœƒ/sin πœƒ = 0 [B] lim_(πœƒ β†’ 0) πœƒ/sin πœƒ does not exist [C] lim_(πœƒ β†’ 0) πœƒ/sin πœƒ = 1

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Video Transcript

Consider the following arc of a unit circle, where ray 𝑂𝑃 is inclined at πœƒ radians. Part one) What, in terms of πœƒ, are the coordinates of 𝑃?

Well, here is 𝑃. And if we take the angle 𝑂𝑇𝑃 to be a right angle, we can see that 𝑃 is directly above the point 𝑇 with coordinates one, zero. And so, the π‘₯-coordinate of 𝑃 must be one. What about the 𝑦-coordinate though? Well, the 𝑦-coordinate of 𝑃 is just the length 𝑇𝑃. We find this length by considering the right triangle 𝑂𝑇𝑃. The origin 𝑂 has coordinates zero, zero. And so, we can see that the length 𝑂𝑇 is just one.

So we know the length of the side adjacent to the angle πœƒ. And we’d like to know the length of the side opposite it. Well then, we should use tan πœƒ, which is the ratio of the length of the side opposite the angle πœƒ. That’s 𝑇𝑃 divided by the length of the side adjacent which is side 𝑂𝑇, which we know has length one. 𝑇𝑃 which remember is the 𝑦-coordinate of 𝑃 is, therefore, tan πœƒ. And so, the coordinates of 𝑃 are one, tan πœƒ.

Part two) Write the following inequalities in terms of sin πœƒ, πœƒ, and cos πœƒ. The length of 𝑄𝑅 is less than the length of the arc from 𝑇 to 𝑄, which is less than the length 𝑇𝑃.

Let’s start with 𝑄𝑅. Again, it’s best to find this by considering a right triangle, in this case the right triangle 𝑂𝑅𝑄. Now, as 𝑄 lies on the unit circle, its distance from the origin is one. And so, 𝑂𝑄 is one. We have the length of the hypotenuse of the right triangle. And we want to find the length of the side opposite to the angle πœƒ. That’s the length 𝑄𝑅. So as we’d like to find the opposite side and we have the hypotenuse, we use sine. Sin of πœƒ is opposite β€” that’s 𝑄𝑅 β€” over hypotenuse one. And so, 𝑄𝑅 is sin πœƒ.

We now find the length of the arc from 𝑇 to 𝑄. That’s this length here. Well, we can see that this arc subtends an angle of πœƒ at the centre of this unit circle. And πœƒ we’re told in the question is measured in radians. By definition then, the length of the arc from 𝑇 to 𝑄 is πœƒ. Alternatively, you manage to find πœƒ in radians as the ratio of the arc length to the radius. But as we’re in the unit circle, the radius is one. And so, πœƒ is just the arc length 𝑇𝑄. What about the length 𝑇𝑃?

Well, if you remember in the first part of the question, we found this to be tan πœƒ. 𝑇𝑃 was the 𝑦-coordinate of the point 𝑃. But we’re asked to give our answer in terms of sin πœƒ, πœƒ, and cos πœƒ alone. So we rewrite tan πœƒ as sin πœƒ over cos πœƒ. Our inequality becomes that sin πœƒ is less than πœƒ which is less than sin πœƒ over cos πœƒ.

Finally, part three) by dividing your inequalities by sin πœƒ, using the squeeze theorem and the fact that the limit of cos πœƒ as πœƒ approaches zero is one, which of the following conclusions can you draw? Is it that A) the limit of πœƒ over sin πœƒ as πœƒ approaches zero is zero, B) that this limit does not exist, or C) that this limit has value one.

We start by dividing the inequalities obtained in the second part of the question by sin πœƒ. Sin πœƒ divided by sin πœƒ is one. πœƒ divided by sin πœƒ can’t be simplified further. But sin πœƒ over cos πœƒ divided by sin πœƒ can. We get one over cos πœƒ. Now, we use the squeeze theorem. But how should we use it? Well, the options give us some clue. Basically, we’re asked for the limits of πœƒ over sin πœƒ as πœƒ approaches zero. So what is this limit? Well, we’ve got two functions either side of πœƒ over sin πœƒ. The limit of one is of course just one. The limit of one over cos πœƒ is slightly more tricky. We have to use the fact that the limit of a quotient is the quotient of the limit.

But having done this, we know what the limit of cos πœƒ as πœƒ approaches zero is. We’re told that in the question. This value is one. And so, the limit of one over cos πœƒ as πœƒ approaches zero is one over one which is one. So both the function one and the function one over cos πœƒ have the same limit: the limit one as πœƒ approaches zero. And so, as πœƒ over sin πœƒ lies between these two functions for values of πœƒ near to zero, we can apply the squeeze theorem. The limit of πœƒ over sin πœƒ as πœƒ approaches zero must be one as well, the same as the other two limits. This is our answer which corresponds to option C.

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