### Video Transcript

In this video, weβre 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 canβt be evaluated using more basic techniques. Letβs start by trying to understand
the statement of the theorem.

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

Hereβs a sketch of the graphs of
two functions π and β, which have the same limits as π₯ approaches π is required
and also π of π₯ is always less than or equal to β of π₯ for any value of π₯ near
π as required. But we still need to sketch the
graph of π of π₯, whose value is always between those of π of π₯ and β of π₯. Letβs try sketching. As π of π₯ lies between π of π₯
and β of π₯, the graph of π of π₯ must lie between the graph of π of π₯ and the
graph of β of π₯. We start off with quite a lot of
room between the graphs of π of π₯ and β of π₯. But as π₯ approaches π and the
graphs of π of π₯ and β of π₯ come together, weβre 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 π. That the limit of π of π₯ as π₯
approaches π must also be πΏ. We have lots of choice over the
value of π of π₯ when π₯ is not near π. But as π₯ approaches π, the value
of π of π₯ must approach πΏ. The squeeze theorem is a very
general theorem. We donβt 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 π of π₯ is less than or equal to π of π₯
which is less than or equal to β of π₯ when π₯ is exactly π, as long as this
inequality holds elsewhere near π.

Hopefully, we understand the
statement of the squeeze theorem better than at the start of the video. Weβre 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 π΄ and π΅ with π΄ of π₯ less than or equal to π΅ of π₯ for π₯ between two
and 3.8. Looking at the diagram then, we see
that this must be π΄ of π₯ and this must be π΅ of π₯.

Weβre asked what does the squeeze
theorem tell us about a continuous function π 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 π of π₯ is less
than or equal to π of π₯ which is less than or equal to β of π₯ when π₯ is near π
and the limit of π of π₯ as π₯ approaches π equals the limit of β of π₯ as π₯
approaches π which equals πΏ. Then the limit of π of π₯ as π₯
approaches π also equals πΏ. How does this help us with our
question?

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

Looking at the graph, we can see
when the limits of π΄ of π₯ and π΅ of π₯ are equal. Itβs as π₯ approaches three when
the limit is one. So this condition is satisfied as
well with π equal to three and πΏ equal to one. And so, we can conclude that the
limit of π of π₯ as π₯ approaches π equals πΏ. Using the fact that π is three and
πΏ is one, we get that the limit of π of π₯ as π₯ approaches three is one. This is our answer. This is what the squeeze theorem
tells us about the continuous function π 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 π of π₯ as π₯ approaches three should be one. In fact, as weβre told that π is a
continuous function, we can go further and say that the value of π of three as it
equals the limit of π of π₯ as π₯ approaches three must be one. Letβs 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 ππ is inclined at π radians. Part one) What, in terms of π, are
the coordinates of π?

Well, here is π. And if we take the angle πππ to
be a right angle, we can see that π is directly above the point π with coordinates
one, zero. And so, the π₯-coordinate of π
must be one. What about the π¦-coordinate
though? Well, the π¦-coordinate of π is
just the length ππ. We find this length by considering
the right triangle πππ. The origin π has coordinates zero,
zero. And so, we can see that the length
ππ is just one.

So we know the length of the side
adjacent to the angle π. And weβd like to know the length of
the side opposite it. Well then, we should use tan π,
which is the ratio of the length of the side opposite the angle π. Thatβs ππ divided by the length
of the side adjacent which is side ππ, which we know has length one. ππ which remember is the
π¦-coordinate of π is, therefore, tan π. And so, the coordinates of π are
one, tan π.

Part two) Write the following
inequalities in terms of sin π, π, and cos π. The length of ππ
is less than the
length of the arc from π to π, which is less than the length ππ.

Letβs start with ππ
. Again, itβs best to find this by
considering a right triangle, in this case the right triangle ππ
π. Now, as π lies on the unit circle,
its distance from the origin is one. And so, ππ 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. Thatβs the length ππ
. So as weβd like to find the
opposite side and we have the hypotenuse, we use sine. Sin of π is opposite β thatβs ππ
β over hypotenuse one. And so, ππ
is sin π.

We now find the length of the arc
from π to π. Thatβs this length here. Well, we can see that this arc
subtends an angle of π at the centre of this unit circle. And π weβre told in the question
is measured in radians. By definition then, the length of
the arc from π to π is π. Alternatively, you manage to find
π in radians as the ratio of the arc length to the radius. But as weβre in the unit circle,
the radius is one. And so, π is just the arc length
ππ. What about the length ππ?

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

Finally, part three) by dividing
your inequalities by sin π, using the squeeze theorem and the fact that the limit
of cos π as π approaches zero is one, which of the following conclusions can you
draw? Is it that A) the limit of π over
sin π as π 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. Sin π divided by sin π is
one. π divided by sin π canβt be
simplified further. But sin π over cos π divided by
sin π can. We get one over cos π. Now, we use the squeeze
theorem. But how should we use it? Well, the options give us some
clue. Basically, weβre asked for the
limits of π over sin π as π approaches zero. So what is this limit? Well, weβve got two functions
either side of π over sin π. The limit of one is of course just
one. The limit of one over cos π 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 as π approaches zero is. Weβre told that in the
question. This value is one. And so, the limit of one over cos
π as π approaches zero is one over one which is one. So both the function one and the
function one over cos π have the same limit: the limit one as π approaches
zero. And so, as π over sin π lies
between these two functions for values of π near to zero, we can apply the squeeze
theorem. The limit of π over sin π as π
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 π over π as π 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 weβre measuring π in radians. If we measured π 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. Letβs now look at our final
example.

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

Letβs remind ourselves of the
squeeze theorem. It says that if π of π₯ is less
than or equal to π of π₯ which is less than or equal to β of π₯ when π₯ is near π
and the limit of π of π₯ equals the limit of β of π₯ as π₯ approaches π. Weβll call this limit πΏ. Then the limit of π of π₯ as π₯
approaches π is also πΏ. Now how do we use this theorem to
calculate this limit here? Well, if we let π of π₯ equal π₯
cubed times cos of two over π₯ and we let π equal zero, then the limit we have to
find is the limit of π of π₯ as π₯ approaches π.

Now if we can find functions π and
β which underestimate and overestimate the function π, respectively, and which have
the same limit as π₯ approaches π which is zero. Then the limit weβre 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 π of π₯ and an overestimate β of π₯ for our function which have the
same limit as π₯ approaches zero. How are we going to do this?

Well, the difficult part of our
function is the factor cos two over π₯. This is the bit of the function
that means that we canβt just do direct substitution. In fact, the limit of cos of two
over π₯ as π₯ 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 π₯. Another way of saying this is that
the absolute value of cosine of two over π₯ is always less than or equal to one. As a result, the absolute value of
our function π of π₯ is less than or equal to the absolute value of π₯ cubed. Another way of saying this is that
π₯ cubed cos two over π₯ is between negative the absolute value of π₯ cubed and the
absolute value of π₯ cubed.

Have we found our functions π of
π₯ and β of π₯ then? Well, maybe. But we need to check that their
limits as π₯ approaches zero are the same. Letβs clear some room to do
this. We need to find the limit of π of
π₯ β thatβs negative the absolute value of π₯ cubed as π₯ approaches zero β and the
limit of β of π₯. Thatβs the absolute value of π₯
cubed as π₯ 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 π of π₯ and
β of π₯ as π₯ approaches zero are the same.

The limit value πΏ is zero. And so, by the squeeze theorem, the
limit of π₯ cubed times cos of two over π₯ as π₯ approaches zero is also zero. This is our answer. Looking at this diagram, you can
see how the graph of the function π₯ cubed times cos of two over π₯ is squeezed
between the graphs of the absolute value of π₯ cubed and itβs opposite. And so, its limit as π₯ approaches
zero must be zero, although the function itself is not defined at π₯ equals
zero. Okay, so letβs recap what weβve
learnt in this video.

Weβve seen the statement of the
theorem if π of π₯ is less than or equal to π of π₯ which is in turn less than or
equal to β of π₯ when π₯ is near π and the limit of π of π₯ as π₯ approaches π
equals the limit of β π₯ as π₯ approaches π which is πΏ. Then the limit of π of π₯ as π₯
approaches π is also πΏ. We can evaluate the limit of π of
π₯ as π₯ approaches π by finding functions π of π₯ and β of π₯ such that π of π₯
is less than or equal to π of π₯ which is less than or equal to β of π₯ near π and
the limit of π of π₯ as π₯ approaches π equals the limit of β of π₯ as π₯
approaches π. Weβll call this πΏ and applying the
squeeze theorem. Upon applying the squeeze theorem,
we find that the limit of π of π₯ as π₯ approaches π is also πΏ. The above method allows us to find
limits of functions which combine polynomials, trigonometric functions, and
quotients. Some of these limits are very
important.