Video: The Squeeze Theorem

In this video, we will learn how to use the squeeze (sandwich) theorem to evaluate some limits when the value of a function is bounded by the values of two other functions.

15:53

Video Transcript

In this video, we’re going to learn about the squeeze theorem. The squeeze theorem is a really useful and important theorem because it helps us to evaluate really useful and important limits that can’t be evaluated using more basic techniques. Let’s start by trying to understand the statement of the theorem.

If 𝑓 of 𝑥 is less than or equal to 𝑔 of 𝑥 which is in turn less than or equal to ℎ of 𝑥 when 𝑥 is near a value 𝑎 and a limit of 𝑓 of 𝑥 as 𝑥 approaches 𝑎 is the same as the limit of ℎ of 𝑥 as 𝑥 approaches 𝑎. We call this limit 𝐿. Then the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎 will also be 𝐿. This is best understood using a diagram. All the action occurs when 𝑥 is near a certain value 𝑎. So let’s mark this special value. Near 𝑎, in other words in some interval around 𝑎, we’re told that 𝑓 of 𝑥 is less than or equal to 𝑔 of 𝑥 which is less than or equal to ℎ of 𝑥. But before we start sketching curves, we should bear in mind the fact that the limit of 𝑓 of 𝑥 as 𝑥 approaches 𝑎 is the same as the limit of ℎ of 𝑥 as 𝑥 approaches 𝑎.

Here’s a sketch of the graphs of two functions 𝑓 and ℎ, which have the same limits as 𝑥 approaches 𝑎 is required and also 𝑓 of 𝑥 is always less than or equal to ℎ of 𝑥 for any value of 𝑥 near 𝑎 as required. But we still need to sketch the graph of 𝑔 of 𝑥, whose value is always between those of 𝑓 of 𝑥 and ℎ of 𝑥. Let’s try sketching. As 𝑔 of 𝑥 lies between 𝑓 of 𝑥 and ℎ of 𝑥, the graph of 𝑔 of 𝑥 must lie between the graph of 𝑓 of 𝑥 and the graph of ℎ of 𝑥. We start off with quite a lot of room between the graphs of 𝑓 of 𝑥 and ℎ of 𝑥. But as 𝑥 approaches 𝑎 and the graphs of 𝑓 of 𝑥 and ℎ of 𝑥 come together, we’re squeezed through this point where they meet. Of course, after this point, we then start to have some room again. This is the squeezing referred to in the name of the squeeze theorem.

And the squeeze theorem states just what we found by trying to sketch a graph of the suitable function 𝑔. That the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎 must also be 𝐿. We have lots of choice over the value of 𝑔 of 𝑥 when 𝑥 is not near 𝑎. But as 𝑥 approaches 𝑎, the value of 𝑔 of 𝑥 must approach 𝐿. The squeeze theorem is a very general theorem. We don’t require that the functions involved are continuous. It applies just as well to functions which are not continuous. And we can make this statement slightly stronger by not requiring that 𝑓 of 𝑥 is less than or equal to 𝑔 of 𝑥 which is less than or equal to ℎ of 𝑥 when 𝑥 is exactly 𝑎, as long as this inequality holds elsewhere near 𝑎.

Hopefully, we understand the statement of the squeeze theorem better than at the start of the video. We’re not going to prove the squeeze theorem in this video. What we are going to do is apply it to various problems to see why it is useful and to strengthen our understanding. The figure shows the graphs of functions 𝐴 and 𝐵 with 𝐴 of 𝑥 less than or equal to 𝐵 of 𝑥 for 𝑥 between two and 3.8. Looking at the diagram then, we see that this must be 𝐴 of 𝑥 and this must be 𝐵 of 𝑥.

We’re asked what does the squeeze theorem tell us about a continuous function 𝑓 whose graph lies in the shaded region over the interval from two to 3.8. Well, what is the squeeze theorem? It says that if 𝑓 of 𝑥 is less than or equal to 𝑔 of 𝑥 which is less than or equal to ℎ of 𝑥 when 𝑥 is near 𝑎 and the limit of 𝑓 of 𝑥 as 𝑥 approaches 𝑎 equals the limit of ℎ of 𝑥 as 𝑥 approaches 𝑎 which equals 𝐿. Then the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎 also equals 𝐿. How does this help us with our question?

Well, we’re told that the graph of 𝑓 lies in the shaded region. That’s between the graphs of 𝐴 of 𝑥 and 𝐵 of 𝑥. And so, for all values of 𝑥 in 𝐵 of 𝑥 open interval from two to 3.8, the value of 𝑓 of 𝑥 must lie between that of 𝐴 of 𝑥 and 𝐵 of 𝑥. We can change the names of the functions in the squeeze theorem. Doing so, we do then indeed have 𝐴 of 𝑥 is less than or equal to 𝑓 of 𝑥 is less than or equal to 𝐵 of 𝑥 when 𝑥 is near 𝐴 or at least in some interval. Now, we also need the limits of 𝐴 of 𝑥 and 𝐵 of 𝑥 as 𝑥 approaches lowercase 𝑎 to be the same.

Looking at the graph, we can see when the limits of 𝐴 of 𝑥 and 𝐵 of 𝑥 are equal. It’s as 𝑥 approaches three when the limit is one. So this condition is satisfied as well with 𝑎 equal to three and 𝐿 equal to one. And so, we can conclude that the limit of 𝑓 of 𝑥 as 𝑥 approaches 𝑎 equals 𝐿. Using the fact that 𝑎 is three and 𝐿 is one, we get that the limit of 𝑓 of 𝑥 as 𝑥 approaches three is one. This is our answer. This is what the squeeze theorem tells us about the continuous function 𝑓 whose graph lies in the shaded region.

This result should make sense. If you try to draw any graph in the shaded region, you will end up passing through this point three, one. It, therefore, makes sense that the limit of 𝑓 of 𝑥 as 𝑥 approaches three should be one. In fact, as we’re told that 𝑓 is a continuous function, we can go further and say that the value of 𝑓 of three as it equals the limit of 𝑓 of 𝑥 as 𝑥 approaches three must be one. Let’s now see an example where the squeeze theorem helps us to evaluate an important limit.

Consider the following arc of a unit circle, where ray 𝑂𝑃 is inclined at 𝜃 radians. Part one) What, in terms of 𝜃, 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 𝑥-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 𝜃. And we’d like to know the length of the side opposite it. Well then, we should use tan 𝜃, which is the ratio of the length of the side opposite the angle 𝜃. 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 𝜃. And so, the coordinates of 𝑃 are one, tan 𝜃.

Part two) Write the following inequalities in terms of sin 𝜃, 𝜃, and cos 𝜃. 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 𝜃. 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 𝜃 is opposite — that’s 𝑄𝑅 — over hypotenuse one. And so, 𝑄𝑅 is sin 𝜃.

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 𝜃 at the centre of this unit circle. And 𝜃 we’re told in the question is measured in radians. By definition then, the length of the arc from 𝑇 to 𝑄 is 𝜃. Alternatively, you manage to find 𝜃 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, 𝜃 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 𝜃. 𝑇𝑃 was the 𝑦-coordinate of the point 𝑃. But we’re asked to give our answer in terms of sin 𝜃, 𝜃, and cos 𝜃 alone. So we rewrite tan 𝜃 as sin 𝜃 over cos 𝜃. Our inequality becomes that sin 𝜃 is less than 𝜃 which is less than sin 𝜃 over cos 𝜃.

Finally, part three) by dividing your inequalities by sin 𝜃, using the squeeze theorem and the fact that the limit of cos 𝜃 as 𝜃 approaches zero is one, which of the following conclusions can you draw? Is it that A) the limit of 𝜃 over sin 𝜃 as 𝜃 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 𝜃. Sin 𝜃 divided by sin 𝜃 is one. 𝜃 divided by sin 𝜃 can’t be simplified further. But sin 𝜃 over cos 𝜃 divided by sin 𝜃 can. We get one over cos 𝜃. 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 𝜃 over sin 𝜃 as 𝜃 approaches zero. So what is this limit? Well, we’ve got two functions either side of 𝜃 over sin 𝜃. The limit of one is of course just one. The limit of one over cos 𝜃 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 𝜃 as 𝜃 approaches zero is. We’re told that in the question. This value is one. And so, the limit of one over cos 𝜃 as 𝜃 approaches zero is one over one which is one. So both the function one and the function one over cos 𝜃 have the same limit: the limit one as 𝜃 approaches zero. And so, as 𝜃 over sin 𝜃 lies between these two functions for values of 𝜃 near to zero, we can apply the squeeze theorem. The limit of 𝜃 over sin 𝜃 as 𝜃 approaches zero must be one as well, the same as the other two limits. This is our answer which corresponds to option C.

As a corollary, we can take the reciprocal and say that the limit of sin 𝜃 over 𝜃 as 𝜃 approaches zero is one. This is a really important limit to know. Because along with the corresponding limit for cosine, it allows us to differentiate trigonometric functions. The fact that we get such a nice simple answer might be surprising. In fact, we only get the value one because we’re measuring 𝜃 in radians. If we measured 𝜃 in degrees, then our answer would be 𝜋 divided by 180, much less elegant. The reason that we use radians rather than some other units of angle is to make this limit one. Let’s now look at our final example.

Calculate the limit of 𝑥 squared times cos of two over 𝑥 as 𝑥 approaches zero using the squeeze theorem.

Let’s remind ourselves of the squeeze theorem. It says that if 𝑓 of 𝑥 is less than or equal to 𝑔 of 𝑥 which is less than or equal to ℎ of 𝑥 when 𝑥 is near 𝑎 and the limit of 𝑓 of 𝑥 equals the limit of ℎ of 𝑥 as 𝑥 approaches 𝑎. We’ll call this limit 𝐿. Then the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎 is also 𝐿. Now how do we use this theorem to calculate this limit here? Well, if we let 𝑔 of 𝑥 equal 𝑥 cubed times cos of two over 𝑥 and we let 𝑎 equal zero, then the limit we have to find is the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎.

Now if we can find functions 𝑓 and ℎ which underestimate and overestimate the function 𝑔, respectively, and which have the same limit as 𝑥 approaches 𝑎 which is zero. Then the limit we’re looking for will also have this value. We write this idea down to remind ourselves of it. Now all we have to do is find an underestimate 𝑓 of 𝑥 and an overestimate ℎ of 𝑥 for our function which have the same limit as 𝑥 approaches zero. How are we going to do this?

Well, the difficult part of our function is the factor cos two over 𝑥. This is the bit of the function that means that we can’t just do direct substitution. In fact, the limit of cos of two over 𝑥 as 𝑥 approaches zero is undefined you can see this by graphing the function using a computer or graphing calculator. However, there are no asymptotes here. The range of cosine is from negative one to one. The cosine of any number will have to lie between negative one and one. And that remains true even that number is two over 𝑥. Another way of saying this is that the absolute value of cosine of two over 𝑥 is always less than or equal to one. As a result, the absolute value of our function 𝑔 of 𝑥 is less than or equal to the absolute value of 𝑥 cubed. Another way of saying this is that 𝑥 cubed cos two over 𝑥 is between negative the absolute value of 𝑥 cubed and the absolute value of 𝑥 cubed.

Have we found our functions 𝑓 of 𝑥 and ℎ of 𝑥 then? Well, maybe. But we need to check that their limits as 𝑥 approaches zero are the same. Let’s clear some room to do this. We need to find the limit of 𝑓 of 𝑥 — that’s negative the absolute value of 𝑥 cubed as 𝑥 approaches zero — and the limit of ℎ of 𝑥. That’s the absolute value of 𝑥 cubed as 𝑥 approaches zero. Both of these functions are continuous. And so, these limits can be evaluated using direct substitution. Negative the absolute value of zero cubed is just zero as is the absolute value of zero cubed. So yes, the limits of 𝑓 of 𝑥 and ℎ of 𝑥 as 𝑥 approaches zero are the same.

The limit value 𝐿 is zero. And so, by the squeeze theorem, the limit of 𝑥 cubed times cos of two over 𝑥 as 𝑥 approaches zero is also zero. This is our answer. Looking at this diagram, you can see how the graph of the function 𝑥 cubed times cos of two over 𝑥 is squeezed between the graphs of the absolute value of 𝑥 cubed and it’s opposite. And so, its limit as 𝑥 approaches zero must be zero, although the function itself is not defined at 𝑥 equals zero.

Okay, so let’s recap what we’ve learnt in this video. We’ve seen the statement of the theorem if 𝑓 of 𝑥 is less than or equal to 𝑔 of 𝑥 which is in turn less than or equal to ℎ of 𝑥 when 𝑥 is near 𝑎 and the limit of 𝑓 of 𝑥 as 𝑥 approaches 𝑎 equals the limit of ℎ 𝑥 as 𝑥 approaches 𝑎 which is 𝐿. Then the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎 is also 𝐿. We can evaluate the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎 by finding functions 𝑓 of 𝑥 and ℎ of 𝑥 such that 𝑓 of 𝑥 is less than or equal to 𝑔 of 𝑥 which is less than or equal to ℎ of 𝑥 near 𝑎 and the limit of 𝑓 of 𝑥 as 𝑥 approaches 𝑎 equals the limit of ℎ of 𝑥 as 𝑥 approaches 𝑎. We’ll call this 𝐿 and applying the squeeze theorem. Upon applying the squeeze theorem, we find that the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎 is also 𝐿. The above method allows us to find limits of functions which combine polynomials, trigonometric functions, and quotients. Some of these limits are very important.

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