### Video Transcript

In this video, we will learn how to
find the limits of trigonometric functions. We will be using some rules to help
us. Letβs start by reminding ourselves
what a limit is. If the limit of π of π₯ as π₯
approaches π exists, then we can say that itβs equal to some constant πΏ. And what this means is that π₯
approaches π. The function π of π₯ approaches
πΏ.

When finding the limits of
trigonometric functions, there are some functions which can be found using direct
substitution, for example, the limits of sin π₯ and cos π₯. Some functions require the use of
trigonometric identities, such as the one shown here, in order to manipulate them
into a form where we can use direct substitution.

However, there are some cases where
direct substitution does not work. One such case is the limit as π₯
approaches zero of sin π₯ over π₯. Since when we try to use direct
substitution here, we obtain sin of zero over zero, which is also equal to zero over
zero. And this is undefined. And so here weβve reached the first
rule which weβll be using to help us find the limits of trigonometric functions. We have that the limit as π₯
approaches zero of sin π₯ over π₯ is equal to one.

Now the proof of this rule is a bit
beyond the scope of this video. However, if we think about this
rule in a certain way, we can gain an intuitive insight into why it works. If we think about our small angle
approximations, we know that when π₯ is very very small, sin of π₯ is roughly equal
to π₯. In our limit, weβre taking the
limit as π₯ approaches zero. So this means π₯ will be getting
smaller and smaller. And therefore, it makes sense for
us to use our small angle approximation here. We obtain that the limit as π₯
approaches zero of sin π₯ over π₯ is roughly equal to π₯ over π₯. Canceling the π₯s on the top and
bottom of the fraction leaves us with one.

Another way to think about this
limit intuitively is to draw the graph of sin π₯ over π₯. We can see from plotting the graph
that the line tends toward one at zero. We would obtain a similar result by
considering a table of values for sin of π₯ over π₯. Now letβs consider an example using
this identity.

Evaluate the limit as π₯ approaches
zero of sin of π₯ over sin of π₯ over two.

First, we can try to solve this
limit by using direct substitution. We substitute π₯ equals zero into
our equation. However, this leaves us with zero
over zero, which is undefined. We will need to find this limit by
another means. Letβs try using the rule that the
limit as π₯ approaches zero of sin π₯ over π₯ is equal to one.

In order to obtain something of
this form, we need to multiply both the numerator and denominator of our limit by
π₯. In doing this, it enables us to
write our limit as the limit as π₯ approaches zero of sin π₯ over π₯ multiplied by
π₯ over sin of π₯ over two.

Next, weβll be using the limit
rule, which tells us that the limit of a product of functions is equal to the
product of the limits of the functions. We obtain this. We notice that the limit on the
left of the product is identical to the limit in our rule. And so we can say that this limit
is just one.

In order to evaluate the other
limit, letβs rewrite our rule that, instead of writing π₯, weβll write π₯ over
two. π₯ over two approaching zero is the
same as π₯ approaching zero. And so we can write this in
here. Next, we can multiply both the
numerator and denominator by two. This leaves us with the limit as π₯
approaches zero of two sin of π₯ over two over π₯ is equal to one.

And now we have a constant inside
our limit, which is two. So we can factorize this constant
out of our limit. And we simply divide both sides of
the equation by two. The limit on the left of the
equation here is looking very close to the limit weβre trying to evaluate. The only difference is that the
fractions in the two limits are reciprocals of one another.

In order to make these two limits
identical, weβll be using the fact that the limit of a reciprocal is equal to the
reciprocal of the limit. What this means is that the limit
as π₯ approaches zero of π₯ over sin of π₯ over two is equal to one over the limit
as π₯ approaches zero of sin of π₯ over two over π₯. We have just found this limit to be
equal to one-half. So we can substitute this in
here. And this gives us one over
one-half, which is simply equal to two.

And so here we have found the value
of the limit which weβre trying to evaluate. And we can substitute this back
into our equation. This tells us that the limit as π₯
approaches zero of sin π₯ over sin of π₯ over two is equal to one multiplied by two,
giving us a solution of two.

An alternative method to solve this
question is to use a trigonometric identity. Weβll be using the fact that sin of
two π is equal to two sin π cos π. If we let π equal π₯ over two, we
obtain that the sin of π₯ is equal to two sin of π₯ over two timesed by cos of π₯
over two. And we can substitute this value of
sin of π₯ into the numerator of our limit. In doing this, we obtain the limit
as π₯ approaches zero of two sin of π₯ over two multiplied by cos of π₯ over two all
over sin of π₯ over two.

And so we can cancel the sin of π₯
over two in the top and bottom of the fraction. This leaves us with the limit as π₯
approaches zero of two cos of π₯ over two. And here we can simply apply direct
substitution. And since cos of zero is equal to
one, we obtain the same solution as earlier of two.

In this last example, we saw how we
can adapt the limit as π₯ approaches zero of sin of π₯ over π₯ is equal to one in
order to show that the limit as π₯ approaches zero of π₯ over sin of π₯ over two is
equal to two. Letβs consider the general case of
the limit as π₯ approaches zero of sin of ππ₯ over π₯.

Taking our original rule and
substituting in ππ₯ for π₯, we obtain this. However, since π is a constant, if
ππ₯ is approaching zero, then this must mean that π₯ is approaching zero. And so instead of writing ππ₯ is
approaching zero, we can simply write π₯ is approaching zero since these two things
are equivalent.

Next, we can factorize out the π
in the denominator of our fraction. Our fraction becomes one over π
multiplied by sin of ππ₯ over π₯. Since one over π is just a
constant, we can factorize it out of our limit.

For our final step here, we
multiply both sides by π. And this leaves us with a new
rule. We have that the limit as π₯
approaches zero of sin of ππ₯ over π₯ is equal to π. We can further adapt this rule in
order to find the limit as π₯ approaches zero of tan of ππ₯ over π₯. We start by writing tan of ππ₯ as
sin of ππ₯ over cos of ππ₯.

Next, weβll be using the fact that
the limit of a product of functions is equal to the product of the limit of those
functions. And we obtain this. And we can see that the limit on
the left-hand side is equivalent to the rule we just derived. And so, therefore, itβs equal to
π. And we can use direct substitution
in order to find the limit on the right-hand side. Since cos of zero is just one, we
obtain that this is equal to π. And so we found a new rule. Thatβs the limit as π₯ approaches
zero of tan ππ₯ over π₯ is equal to π. Now weβre ready to move on to our
next example.

Find the limit as π₯ approaches
zero of sin squared of seven π₯ plus three tan squared of three π₯ over eight π₯
squared.

If we were to try direct
substitution, we would obtain zero over zero, which is undefined. Letβs try to find this limit using
these rules. Weβll also be using the fact that
the limit of a sum of functions is equal to the sum of the limits of the
functions. Hence, we can write our limit as
the sum of these two limits. We notice that we can take a factor
of one-eighth out of the first limit and a factor of three-eighths out of the second
limit.

Next, we notice that both
numerators and both denominators are squares, enabling us to write our limits like
this. Now we can use the fact that the
limit of a square of a function is equal to the square of the limit of the
function. In doing this, we are left with
this. And we notice that our limits look
very similar to the ones we wrote out at the start.

Substituting π equals seven into
the first limit rule, we see that our limit on the left must be equal to seven. And substituting π equals three
into the second limit rule, we see that our limit on the right must be equal to
three. We obtain one-eighth multiplied by
seven squared plus three-eighths multiplied by three squared. Simplifying this down, we obtain a
solution of 19 over two.

Next, weβll be looking at a
different rule which is useful for finding the limits of trigonometric functions of
another form. The rule which weβll be using is
the limit as π₯ approaches zero of one minus cos π₯ over π₯ is equal to zero. Now the proof of this rule is again
beyond the scope of this video. However, we can think about this
rule intuitively by considering the small angle approximation of cos.

We have that, for small values of
π₯, cos of π₯ is roughly equal to one minus π₯ squared over two. So as π₯ approaches zero, cos of π₯
will be approaching one minus π₯ squared over two. We obtain that the limit as π₯
approaches zero of one minus cos of π₯ over π₯ is roughly equal to the limit as π₯
approaches zero of one minus one minus π₯ squared over two all over π₯, which
simplifies to the limit as π₯ approaches zero of π₯ over two. Using direct substitution, we see
that this is equal to zero, which agrees with the rule which we stated at the
beginning.

Another way to see this intuitively
is to consider the graph of one minus cos of π₯ over π₯. We can see from the graph that as
the value of π₯ approaches zero, the line of the graph also approaches zero. We would see a similar result by
using a table of values. Letβs now consider another
example.

Determine the limit as π₯
approaches zero of nine minus nine cos of seven π₯ over three π₯.

First, we notice that we can cancel
a factor of three from the top and bottom of this fraction, leaving us with this
limit. Next, we notice that we can factor
three out of the limit. If we attempt to do direct
substitution at this point, we will see that we obtain zero over zero, which is
undefined.

Letβs instead use the fact that the
limit as π₯ approaches zero of one minus cos of π₯ over π₯ is equal to zero. We can in fact adapt this formula
in order to find that the limit as π₯ approaches zero of one minus cos of ππ₯ over
π₯ is equal to zero, where π is just a constant. We can do this by substituting ππ₯
in for π₯ into the first rule. We obtain that the limit as ππ₯
goes to zero of one minus cos of ππ₯ over ππ₯ is equal to zero.

Now since π is a constant and ππ₯
is going to zero, this means that π₯ is also going to zero. Instead of writing ππ₯ goes to
zero, we can simply write π₯ goes to zero. Next, we notice we have a factor of
π in the denominator of this fraction. And so we can factorize this π out
of the limit.

Next, we can multiply both sides of
the equation by π. Since the right-hand side is zero,
zero times π gives us zero. So the right-hand side remains as
zero. And so now we have found that the
limit as π₯ goes to zero of one minus cos of ππ₯ over π₯ is zero, which is what we
were trying to show. By substituting π equals seven
into this formula, we can see that our limit here is simply zero. And since three multiplied by zero
is simply zero, we find that the solution here is simply zero.

Next, letβs look at a slightly more
tricky example.

Find the limit as π₯ goes to π
over two of two minus two sin of π₯ over four π₯ minus two π.

First, we try solving it by direct
substitution. However, weβll reach zero over
zero, which is undefined. So we must find this limit by
another means. Letβs start by cancelling a factor
of two in both the numerator and denominator of the fraction. Now letβs consider some of the
rules we know. When considering the first rule
here, we notice that the value inside of the sine must be the same as the value in
the denominator of the function. In our limit, we have sin of π₯ in
the numerator. However, in the denominator, we
have two π₯ minus π. And these two things are not
equal. Therefore, we cannot use this first
rule.

In order to use the second rule, we
require a cos π₯ in the numerator. However, in our limit, we currently
have a sine. Here weβll be using an identity in
order to change the sine to a cosine. We have that sin of π₯ is equal to
cos of π₯ minus π by two. And we can substitute this into our
limit. Factorizing the denominator of the
fraction here, we can see that this is now of a very similar form to the rule which
we know.

At this stage, we need to perform a
substitution. We will substitute in π’ is equal
to π₯ minus π by two. However, we first need to consider
what will happen to our limit. So thatβs π₯ approaching π by
two. Well, weβll simply consider what
happens to the value of π’ as π₯ approaches π by two. Our value of π’ will approach π by
two minus π by two, which is simply zero. And so now weβre ready to
substitute π’ is equal to π₯ minus π by two into our limit. We obtain the limit as π’
approaches zero of one minus cos of π’ over two π’. We can factor out the half. And now we notice that our limit is
identical to our rule. And so, therefore, it must equal
zero. And this gives us a solution of
zero.

In this last example, we saw how we
have to be careful with trigonometric limits as sometimes it can be difficult to
spot how to solve them. Itβs very important to keep the
trigonometric identities in mind.

Now letβs recap some of the key
points of this video. Key Points. The limit as π₯ approaches zero of
sin of π₯ over π₯ is equal to one. This gives us that the limit as π₯
approaches zero of sin of ππ₯ over π₯ is equal to π. And this then leads us on to the
limit as π₯ approaches zero of tan of ππ₯ over π₯ is equal to π. We also have that the limit as π₯
approaches zero of one minus cos of π₯ over π₯ is equal to zero, which leads on to
the limit as π₯ approaches zero of one minus cos of ππ₯ over π₯ is also equal to
zero. And if we cannot solve a limit of a
trigonometric function using direct substitution or one of the rules above, then we
should try using some trigonometric identities, such as the ones shown here.