# Lesson Video: Limits of Trigonometric Functions Mathematics • Higher Education

In this video, we will learn how to evaluate limits of trigonometric functions.

16:09

### 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. Let’s start by reminding ourselves what a limit is. If the limit of 𝑓 of 𝑥 as 𝑥 approaches 𝑎 exists, then we can say that it’s equal to some constant 𝐿. And what this means is that 𝑥 approaches 𝑎. The function 𝑓 of 𝑥 approaches 𝐿.

When finding the limits of trigonometric functions, there are some functions which can be found using direct substitution, for example, the limits of sin 𝑥 and cos 𝑥. 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 𝑥 approaches zero of sin 𝑥 over 𝑥. 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 we’ve reached the first rule which we’ll be using to help us find the limits of trigonometric functions. We have that the limit as 𝑥 approaches zero of sin 𝑥 over 𝑥 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 𝑥 is very very small, sin of 𝑥 is roughly equal to 𝑥. In our limit, we’re taking the limit as 𝑥 approaches zero. So this means 𝑥 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 𝑥 approaches zero of sin 𝑥 over 𝑥 is roughly equal to 𝑥 over 𝑥. Canceling the 𝑥s 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 𝑥 over 𝑥. 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 𝑥 over 𝑥. Now let’s consider an example using this identity.

Evaluate the limit as 𝑥 approaches zero of sin of 𝑥 over sin of 𝑥 over two.

First, we can try to solve this limit by using direct substitution. We substitute 𝑥 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. Let’s try using the rule that the limit as 𝑥 approaches zero of sin 𝑥 over 𝑥 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 𝑥. In doing this, it enables us to write our limit as the limit as 𝑥 approaches zero of sin 𝑥 over 𝑥 multiplied by 𝑥 over sin of 𝑥 over two.

Next, we’ll 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, let’s rewrite our rule that, instead of writing 𝑥, we’ll write 𝑥 over two. 𝑥 over two approaching zero is the same as 𝑥 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 𝑥 approaches zero of two sin of 𝑥 over two over 𝑥 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 we’re 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, we’ll 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 𝑥 approaches zero of 𝑥 over sin of 𝑥 over two is equal to one over the limit as 𝑥 approaches zero of sin of 𝑥 over two over 𝑥. 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 we’re trying to evaluate. And we can substitute this back into our equation. This tells us that the limit as 𝑥 approaches zero of sin 𝑥 over sin of 𝑥 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. We’ll be using the fact that sin of two 𝜃 is equal to two sin 𝜃 cos 𝜃. If we let 𝜃 equal 𝑥 over two, we obtain that the sin of 𝑥 is equal to two sin of 𝑥 over two timesed by cos of 𝑥 over two. And we can substitute this value of sin of 𝑥 into the numerator of our limit. In doing this, we obtain the limit as 𝑥 approaches zero of two sin of 𝑥 over two multiplied by cos of 𝑥 over two all over sin of 𝑥 over two.

And so we can cancel the sin of 𝑥 over two in the top and bottom of the fraction. This leaves us with the limit as 𝑥 approaches zero of two cos of 𝑥 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 𝑥 approaches zero of sin of 𝑥 over 𝑥 is equal to one in order to show that the limit as 𝑥 approaches zero of 𝑥 over sin of 𝑥 over two is equal to two. Let’s consider the general case of the limit as 𝑥 approaches zero of sin of 𝑎𝑥 over 𝑥.

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

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

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

Next, we’ll 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, it’s equal to 𝑎. 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 𝑎. And so we found a new rule. That’s the limit as 𝑥 approaches zero of tan 𝑎𝑥 over 𝑥 is equal to 𝑎. Now we’re ready to move on to our next example.

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

If we were to try direct substitution, we would obtain zero over zero, which is undefined. Let’s try to find this limit using these rules. We’ll 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 𝑎 equals seven into the first limit rule, we see that our limit on the left must be equal to seven. And substituting 𝑎 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, we’ll be looking at a different rule which is useful for finding the limits of trigonometric functions of another form. The rule which we’ll be using is the limit as 𝑥 approaches zero of one minus cos 𝑥 over 𝑥 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 𝑥, cos of 𝑥 is roughly equal to one minus 𝑥 squared over two. So as 𝑥 approaches zero, cos of 𝑥 will be approaching one minus 𝑥 squared over two. We obtain that the limit as 𝑥 approaches zero of one minus cos of 𝑥 over 𝑥 is roughly equal to the limit as 𝑥 approaches zero of one minus one minus 𝑥 squared over two all over 𝑥, which simplifies to the limit as 𝑥 approaches zero of 𝑥 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 𝑥 over 𝑥. We can see from the graph that as the value of 𝑥 approaches zero, the line of the graph also approaches zero. We would see a similar result by using a table of values. Let’s now consider another example.

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

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.

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

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

Next, we can multiply both sides of the equation by 𝑎. Since the right-hand side is zero, zero times 𝑎 gives us zero. So the right-hand side remains as zero. And so now we have found that the limit as 𝑥 goes to zero of one minus cos of 𝑎𝑥 over 𝑥 is zero, which is what we were trying to show. By substituting 𝑎 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, let’s look at a slightly more tricky example.

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

First, we try solving it by direct substitution. However, we’ll reach zero over zero, which is undefined. So we must find this limit by another means. Let’s start by cancelling a factor of two in both the numerator and denominator of the fraction. Now let’s 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 𝑥 in the numerator. However, in the denominator, we have two 𝑥 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 𝑥 in the numerator. However, in our limit, we currently have a sine. Here we’ll be using an identity in order to change the sine to a cosine. We have that sin of 𝑥 is equal to cos of 𝑥 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 𝑢 is equal to 𝑥 minus 𝜋 by two. However, we first need to consider what will happen to our limit. So that’s 𝑥 approaching 𝜋 by two. Well, we’ll simply consider what happens to the value of 𝑢 as 𝑥 approaches 𝜋 by two. Our value of 𝑢 will approach 𝜋 by two minus 𝜋 by two, which is simply zero. And so now we’re ready to substitute 𝑢 is equal to 𝑥 minus 𝜋 by two into our limit. We obtain the limit as 𝑢 approaches zero of one minus cos of 𝑢 over two 𝑢. 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. It’s very important to keep the trigonometric identities in mind.

Now let’s recap some of the key points of this video. Key Points. The limit as 𝑥 approaches zero of sin of 𝑥 over 𝑥 is equal to one. This gives us that the limit as 𝑥 approaches zero of sin of 𝑎𝑥 over 𝑥 is equal to 𝑎. And this then leads us on to the limit as 𝑥 approaches zero of tan of 𝑎𝑥 over 𝑥 is equal to 𝑎. We also have that the limit as 𝑥 approaches zero of one minus cos of 𝑥 over 𝑥 is equal to zero, which leads on to the limit as 𝑥 approaches zero of one minus cos of 𝑎𝑥 over 𝑥 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.