Lesson Explainer: Limits of Trigonometric Functions | Nagwa Lesson Explainer: Limits of Trigonometric Functions | Nagwa

Lesson Explainer: Limits of Trigonometric Functions Mathematics • Second Year of Secondary School

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

Limits are a useful tool for helping us understand the shape of a function around a value; it is one of the fundamental building blocks of calculus. We can find the limit of any trigonometric function by using direct substitution.

Definition: Evaluating the Limit of Trigonometric Functions

If 𝑎 is in the domain of a trigonometric function, then we can evaluate its limit at 𝑎 by direct substitution. In particular, for any 𝑎,

  • limsinsin𝑥=𝑎,
  • limcoscos𝑥=𝑎.

For any 𝑎 in the domain of tan𝑥,

  • limtantan𝑥=𝑎.

These results allow us to evaluate the limit of many trigonometric expressions. However, there are examples that we cannot evaluate. For example, consider limsin𝑥𝑥, where 𝑥 is measured in radians. If we attempt to evaluate this limit using direct substitution, sin00=00, we find an indeterminate form, which means we need to evaluate this limit in a different manner. One way of doing this is to sketch the graph of 𝑦=𝑥𝑥sin.

From the sketch, we can see that as the values of 𝑥 approach 0 from either side, the outputs of the function approach 1. Hence, the sketch indicates that limsin𝑥𝑥=1. We can also see this by constructing a table.

𝑥0.10.010.00100.0010.010.1
sin𝑥𝑥0.998330.999980.999990.999990.999980.99833

Once again, the table suggests that as the values of 𝑥 approach 0 from either side, the outputs of the function approach 1. It is worth noting that we can show a similar result when 𝑥 is measured in degrees; however, when taking limits, we almost always use radians. So, unless otherwise stated, we will assume that the limit of any trigonometric functions involves angles measured in radians. This gives us the following result.

Theorem: Limit of a Trigonometric Expression

If 𝑥 is measured in radians, then limsin𝑥𝑥=1.

We can use this result to show an even more general result. Let 𝑎{0}. We substitute 𝜃=𝑎𝑥 into the limit result limsin𝜃𝜃=1. Note that as 𝜃0, both 𝑎𝑥0 and 𝑥0. This gives us 1=𝜃𝜃=𝑎𝑥𝑎𝑥.limsinlimsin

Taking the factor of 1𝑎 out of this limit and rearranging gives us limsin𝑎𝑥𝑥=𝑎.

It is worth noting this result also holds when 𝑎=0. We can summarize this as follows.

Theorem: Limit of a Trigonometric Expression

If 𝑥 is measured in radians and 𝑎, then limsin𝑎𝑥𝑥=𝑎.

Let’s see an example of using this result to evaluate the limit of a trigonometric expression.

Example 1: Finding Limits Involving Trigonometric Functions

Evaluate limsinsin𝑥.

Answer

Since this limit involves the quotient of trigonometric functions, we can attempt to evaluate this limit by direct substitution sinsin0=00.

This gives us an indeterminate form, which means that we cannot evaluate this limit by direct substitution. Instead, we will use the fact that if 𝑥 is measured in radians and 𝑎 is a real constant, then limsin𝑎𝑥𝑥=𝑎. Although the question does not state that 𝑥 is measured in radians, when taking limits, we almost always work in radians, so we will assume this for the question. We can rewrite the limit as follows: limsinsinlimsinsin𝑥=𝑥𝑥×𝑥.

Assuming that both limits exist, we can write this as the product of two limits:limsinsinlimsinlimsin𝑥𝑥×𝑥=𝑥𝑥×𝑥.

We take the reciprocal of the second limit by using the power rule for limits to get limsinlimsinlimsinlimsin𝑥𝑥×𝑥=𝑥𝑥×𝑥, provided the limit exists and is nonzero. We can then evaluate both of these limits by using our limit result, limsin𝑎𝑥𝑥=𝑎.

The first limit has 𝑎=1 and the second has 𝑎=12. Hence, limsinlimsin𝑥𝑥×𝑥=1×12=2.

There are two more useful limit results involving trigonometric functions that we can find by investigating their graph or by using a table. Consider the following sketches of tan𝑥𝑥 and 1𝑥𝑥cos, where 𝑥 is measured in radians.

In the first diagram, we see that as the values of 𝑥 approach 0, the outputs approach 1. So, the sketch suggests limtan𝑥𝑥=1. Similarly, in the second diagram, as the values of 𝑥 approach 0, we can see that the outputs approach 0. So, the sketch suggests that limcos1𝑥𝑥=0. This gives us the following results.

Theorem: Limit of a Trigonometric Expression

If 𝑥 is measured in radians, then

  • limtan𝑥𝑥=1,
  • limcos1𝑥𝑥=0.

As with the limit result involving sine, we can use substitution to find a limit result where the argument is a constant multiple. If 𝑎, using 𝜃=𝑎𝑥 we have 1=𝜃𝜃=𝑎𝑥𝑎𝑥.limtanlimtan

Taking out the constant factor of 1𝑎 and rearranging gives limtan𝑎𝑥𝑥=𝑎.

Similarly, if 𝑎, using 𝜃=𝑎𝑥 we have 0=1𝜃𝜃=1𝑎𝑥𝑎𝑥.limcoslimcos

Taking out the constant factor of 1𝑎 and rearranging gives limcos1𝑎𝑥𝑥=0.

We can summarize this as follows.

Theorem: Limit of a Trigonometric Expression

If 𝑥 is measured in radians and 𝑎, then

  • limtan𝑎𝑥𝑥=𝑎,
  • limcos1𝑎𝑥𝑥=0.

Let’s see an example of how we can apply these limit results to evaluate the limit of a trigonometric expression.

Example 2: Finding Limits Involving Trigonometric Functions

Determine limcos997𝑥3𝑥.

Answer

Since this limit involves a trigonometric function, we can attempt to evaluate this limit by direct substitution: 99(7×0)3(0)=00.cos

This gives us an indeterminate form, which means we cannot evaluate this limit by direct substitution. Instead, we will use the fact that if 𝑥 is measured in radians and 𝑎, then limcos1𝑎𝑥𝑥=0.

To apply this result, we simplify our limit as follows: limcoslimcoslimcoslimcos997𝑥3𝑥=9(17𝑥)3𝑥=3(17𝑥)𝑥=3×(17𝑥)𝑥=3×0=0.

Hence, limcos997𝑥3𝑥=0.

In our next example, we will use a limit result involving the tangent and sine functions to evaluate the limit of a trigonometric function.

Example 3: Finding Limits Involving Trigonometric Functions

Find limsintan7𝑥+33𝑥8𝑥.

Answer

Since this is the limit of a trigonometric and algebraic expression, we can attempt to evaluate this limit by direct substitution: sintan(7(0))+3(3(0))8(0)=00.

Since this is an indeterminate form, we cannot determine the value of this limit from direct substitution. Instead, we will rewrite this limit in terms of limits we can evaluate. Namely, for any real constant 𝑎 and 𝑥 measured in radians, limtanandlimsin𝑎𝑥𝑥=𝑎𝑎𝑥𝑥=𝑎.

We can rewrite the limit in the question as follows:limsintanlimsintanlimsinlimtanlimsinlimtanlimsinlimtan7𝑥+33𝑥8𝑥=7𝑥8𝑥+33𝑥8𝑥=7𝑥8𝑥+33𝑥8𝑥=187𝑥𝑥+383𝑥𝑥=187𝑥𝑥+383𝑥𝑥.

We can evaluate each of these limits separately. First, we recall that if 𝑥 is measured in radians and constant 𝑎, limsin𝑎𝑥𝑥=𝑎. Using this result, we have limsin7𝑥𝑥=7.

Next, we recall that if 𝑥 is measured in radians and 𝑎, then limtan𝑎𝑥𝑥=𝑎.

Hence, limtan3𝑥𝑥=3.

Substituting the values of these limits into the equation gives us limsintanlimsinlimtan7𝑥+33𝑥8𝑥=187𝑥𝑥+383𝑥𝑥=18(7)+38(3)=498+278=768=192.

Therefore, limsintan7𝑥+33𝑥8𝑥=192.

In our next example, we will combine a trigonometric identity with the limit results of trigonometric functions to evaluate a limit.

Example 4: Finding Limits Involving Trigonometric Functions

Find limsin22𝑥4𝑥2𝜋.

Answer

Since this is the limit of a trigonometric and algebraic expression, we can attempt to evaluate this limit by direct substitution: 2242𝜋=00.sin

Since this is an indeterminate form, we cannot determine the value of this limit from direct substitution. Instead, we will rewrite this limit in terms of limits we can evaluate. We rewrite the limit as follows: limsinlimsinlimsin22𝑥4𝑥2𝜋=2(1𝑥)4𝑥=1𝑥2𝑥.

To evaluate this limit, we will use the substitution 𝜃=𝑥𝜋2. As 𝑥 approaches 𝜋2, 𝜃 will approaches 0. This gives us limsinlimsin1𝑥2𝑥=1𝜃+2𝜃.

Recall that sincos𝜃+𝜋2𝜃. We can use this to rewrite the limit as limsinlimcoslimcos1𝜃+2𝜃=1𝜃2𝜃=121𝜃𝜃.

Finally, we recall that limcos1𝑎𝑥𝑥=0.

Hence, 121𝜃𝜃=12×0=0.limcos

Therefore, limsin22𝑥4𝑥2𝜋=0.

In our final example, we will use these limit results to evaluate the limit of a reciprocal trigonometric expression.

Example 5: Finding Limits Involving Trigonometric Functions

Find limcotcsc6𝑥4𝑥8𝑥.

Answer

Since this is the limit of a trigonometric and algebraic expression, we can attempt to evaluate this limit by direct substitution. However, 0 is not in the domain of this function. Instead, we will rewrite the limit by first using the reciprocal trigonometric identities: limcotcsclimlimsintan6𝑥4𝑥8𝑥=6𝑥=6𝑥8𝑥4𝑥.tansin

We can then rewrite this in terms of the limit results. If 𝑥 is measured in radians and 𝑎, limsinlimtan𝑎𝑥𝑥=𝑎,𝑎𝑥𝑥=𝑎.

So, we have limsintanlimsintanlimtanlimsinlimtanlimsinlimtanlimsinlimtanlimsin6𝑥8𝑥4𝑥=6𝑥8𝑥𝑥4𝑥=6𝑥4𝑥8𝑥𝑥=6𝑥4𝑥8𝑥𝑥=6𝑥4𝑥8𝑥𝑥=64𝑥𝑥8𝑥𝑥.

Applying the limit results, we conclude that 64𝑥𝑥8𝑥𝑥=6(4)(8)=3.limtanlimsin

Hence, limcotcsc6𝑥4𝑥8𝑥=3.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We can evaluate the limit of any trigonometric function at 𝑥=𝑎 by direct substitution if a is in its domain.
  • If 𝑥 is measured in radians, we have the following trigonometric limit results:
    • limsin𝑥𝑥=1,
    • limtan𝑥𝑥=1,
    • limcos1𝑥𝑥=0.
  • If 𝑥 is measured in radians and 𝑎, we have the following trigonometric limit results:
    • limsin𝑎𝑥𝑥=𝑎,
    • limtan𝑎𝑥𝑥=𝑎,
    • limcos1𝑎𝑥𝑥=0.
  • We can use these results to evaluate the limit of trigonometric functions.

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