Lesson Explainer: Limits of Trigonometric Functions Mathematics • Higher Education

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 ๐‘Ž is in the domain of a trigonometric function, then we can evaluate its limit at ๐‘Ž by direct substitution. In particular, for any ๐‘Žโˆˆโ„,

  • limsinsin๏—โ†’๏Œบ๐‘ฅ=๐‘Ž,
  • limcoscos๏—โ†’๏Œบ๐‘ฅ=๐‘Ž.

For any ๐‘Ž in the domain of tan๐‘ฅ,

  • limtantan๏—โ†’๏Œบ๐‘ฅ=๐‘Ž.

These results allow us to evaluate the limit of many trigonometric expressions. However, there are examples that we cannot evaluate. For example, consider limsin๏—โ†’๏Šฆ๐‘ฅ๐‘ฅ, where ๐‘ฅ 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 ๐‘ฆ=๐‘ฅ๐‘ฅsin.

From the sketch, we can see that as the values of ๐‘ฅ approach 0 from either side, the outputs of the function approach 1. Hence, the sketch indicates that limsin๏—โ†’๏Šฆ๐‘ฅ๐‘ฅ=1. We can also see this by constructing a table.

๐‘ฅโˆ’0.1โˆ’0.01โˆ’0.001โ†’0โ†0.0010.010.1
sin๐‘ฅ๐‘ฅ0.998330.999980.99999โ†’โ†0.999990.999980.99833

Once again, the table suggests that as the values of ๐‘ฅ approach 0 from either side, the outputs of the function approach 1. It is worth noting that we can show a similar result when ๐‘ฅ 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 ๐‘ฅ is measured in radians, then limsin๏—โ†’๏Šฆ๐‘ฅ๐‘ฅ=1.

We can use this result to show an even more general result. Let ๐‘Žโˆˆโ„โˆ’{0}. We substitute ๐œƒ=๐‘Ž๐‘ฅ into the limit result limsin๏ผโ†’๏Šฆ๐œƒ๐œƒ=1. Note that as ๐œƒโ†’0, both ๐‘Ž๐‘ฅโ†’0 and ๐‘ฅโ†’0. This gives us 1=๐œƒ๐œƒ=๐‘Ž๐‘ฅ๐‘Ž๐‘ฅ.limsinlimsin๏ผโ†’๏Šฆ๏—โ†’๏Šฆ

Taking the factor of 1๐‘Ž out of this limit and rearranging gives us limsin๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž.

It is worth noting this result also holds when ๐‘Ž=0. We can summarize this as follows.

Theorem: Limit of a Trigonometric Expression

If ๐‘ฅ is measured in radians and ๐‘Žโˆˆโ„, then limsin๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž.

Letโ€™s see an example of using this result to evaluate the limit of a trigonometric expression.

Example 1: Finding Limits Involving Trigonometric Functions

Evaluate limsinsin๏—โ†’๏Šฆ๏—๏Šจ๐‘ฅ๏€ป๏‡.

Answer

Since this limit involves the quotient of trigonometric functions, we can attempt to evaluate this limit by direct substitution sinsin0๏€ป๏‡=00.๏Šฆ๏Šจ

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 ๐‘ฅ is measured in radians and ๐‘Ž is a real constant, then limsin๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž. Although the question does not state that ๐‘ฅ 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: limsinsinlimsinsin๏—โ†’๏Šฆ๏—๏Šจ๏—โ†’๏Šฆ๏—๏Šจ๐‘ฅ๏€ป๏‡=๏‚๐‘ฅ๐‘ฅร—๐‘ฅ๏€ป๏‡๏Ž.

Assuming that both limits exist, we can write this as the product of two limits:limsinsinlimsinlimsin๏—โ†’๏Šฆ๏—๏Šจ๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏—๏Šจ๏‚๐‘ฅ๐‘ฅร—๐‘ฅ๏€ป๏‡๏Ž=๏€ฝ๐‘ฅ๐‘ฅ๏‰ร—๏‚๐‘ฅ๏€ป๏‡๏Ž.

We take the reciprocal of the second limit by using the power rule for limits to get limsinlimsinlimsinlimsin๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏—๏Šจ๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏—๏Šจ๏Šฑ๏Šง๏€ฝ๐‘ฅ๐‘ฅ๏‰ร—๏‚๐‘ฅ๏€ป๏‡๏Ž=๏€ฝ๐‘ฅ๐‘ฅ๏‰ร—๏‚๏‚๏€ป๏‡๐‘ฅ๏Ž๏Ž, provided the limit exists and is nonzero. We can then evaluate both of these limits by using our limit result, limsin๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž.

The first limit has ๐‘Ž=1 and the second has ๐‘Ž=12. Hence, limsinlimsin๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏—๏Šจ๏Šฑ๏Šง๏Šฑ๏Šง๏€ฝ๐‘ฅ๐‘ฅ๏‰ร—๏‚๏‚๏€ป๏‡๐‘ฅ๏Ž๏Ž=1ร—๏€ผ12๏ˆ=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 tan๐‘ฅ๐‘ฅ and 1โˆ’๐‘ฅ๐‘ฅcos, where ๐‘ฅ is measured in radians.

In the first diagram, we see that as the values of ๐‘ฅ approach 0, the outputs approach 1. So, the sketch suggests limtan๏—โ†’๏Šฆ๐‘ฅ๐‘ฅ=1. Similarly, in the second diagram, as the values of ๐‘ฅ approach 0, we can see that the outputs approach 0. So, the sketch suggests that limcos๏—โ†’๏Šฆ1โˆ’๐‘ฅ๐‘ฅ=0. This gives us the following results.

Theorem: Limit of a Trigonometric Expression

If ๐‘ฅ is measured in radians, then

  • limtan๏—โ†’๏Šฆ๐‘ฅ๐‘ฅ=1,
  • limcos๏—โ†’๏Šฆ1โˆ’๐‘ฅ๐‘ฅ=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 ๐‘Žโˆˆโ„, using ๐œƒ=๐‘Ž๐‘ฅ we have 1=๐œƒ๐œƒ=๐‘Ž๐‘ฅ๐‘Ž๐‘ฅ.limtanlimtan๏ผโ†’๏Šฆ๏—โ†’๏Šฆ

Taking out the constant factor of 1๐‘Ž and rearranging gives limtan๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž.

Similarly, if ๐‘Žโˆˆโ„, using ๐œƒ=๐‘Ž๐‘ฅ we have 0=1โˆ’๐œƒ๐œƒ=1โˆ’๐‘Ž๐‘ฅ๐‘Ž๐‘ฅ.limcoslimcos๏ผโ†’๏Šฆ๏—โ†’๏Šฆ

Taking out the constant factor of 1๐‘Ž and rearranging gives limcos๏—โ†’๏Šฆ1โˆ’๐‘Ž๐‘ฅ๐‘ฅ=0.

We can summarize this as follows.

Theorem: Limit of a Trigonometric Expression

If ๐‘ฅ is measured in radians and ๐‘Žโˆˆโ„, then

  • limtan๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž,
  • limcos๏—โ†’๏Šฆ1โˆ’๐‘Ž๐‘ฅ๐‘ฅ=0.

Letโ€™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 limcos๏—โ†’๏Šฆ9โˆ’97๐‘ฅ3๐‘ฅ.

Answer

Since this limit involves a trigonometric function, we can attempt to evaluate this limit by direct substitution: 9โˆ’9(7ร—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 ๐‘ฅ is measured in radians and ๐‘Žโˆˆโ„, then limcos๏—โ†’๏Šฆ1โˆ’๐‘Ž๐‘ฅ๐‘ฅ=0.

To apply this result, we simplify our limit as follows: limcoslimcoslimcoslimcos๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏—โ†’๏Šฆ9โˆ’97๐‘ฅ3๐‘ฅ=9(1โˆ’7๐‘ฅ)3๐‘ฅ=3(1โˆ’7๐‘ฅ)๐‘ฅ=3ร—(1โˆ’7๐‘ฅ)๐‘ฅ=3ร—0=0.

Hence, limcos๏—โ†’๏Šฆ9โˆ’97๐‘ฅ3๐‘ฅ=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 limsintan๏—โ†’๏Šฆ๏Šจ๏Šจ๏Šจ7๐‘ฅ+33๐‘ฅ8๐‘ฅ.

Answer

Since this is the limit of a trigonometric and algebraic expression, we can attempt to evaluate this limit by direct substitution: sintan๏Šจ๏Šจ๏Šจ(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 ๐‘Ž and ๐‘ฅ measured in radians, limtanandlimsin๏—โ†’๏Šฆ๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž.

We can rewrite the limit in the question as follows:limsintanlimsintanlimsinlimtanlimsinlimtanlimsinlimtan๏—โ†’๏Šฆ๏Šจ๏Šจ๏Šจ๏—โ†’๏Šฆ๏Šจ๏Šจ๏Šจ๏Šจ๏—โ†’๏Šฆ๏Šจ๏Šจ๏—โ†’๏Šฆ๏Šจ๏Šจ๏—โ†’๏Šฆ๏Šจ๏Šจ๏—โ†’๏Šฆ๏Šจ๏Šจ๏—โ†’๏Šฆ๏Šจ๏—โ†’๏Šฆ๏Šจ7๐‘ฅ+33๐‘ฅ8๐‘ฅ=๏€พ7๐‘ฅ8๐‘ฅ+33๐‘ฅ8๐‘ฅ๏Š=๏€พ7๐‘ฅ8๐‘ฅ๏Š+๏€พ33๐‘ฅ8๐‘ฅ๏Š=18๏€พ7๐‘ฅ๐‘ฅ๏Š+38๏€พ3๐‘ฅ๐‘ฅ๏Š=18๏€ฝ๏€ฝ7๐‘ฅ๐‘ฅ๏‰๏‰+38๏€ผ๏€ผ3๐‘ฅ๐‘ฅ๏ˆ๏ˆ.

We can evaluate each of these limits separately. First, we recall that if ๐‘ฅ is measured in radians and constant ๐‘Žโˆˆโ„, limsin๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž. Using this result, we have limsin๏—โ†’๏Šฆ๏€ฝ7๐‘ฅ๐‘ฅ๏‰=7.

Next, we recall that if ๐‘ฅ is measured in radians and ๐‘Žโˆˆโ„, then limtan๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž.

Hence, limtan๏—โ†’๏Šฆ๏€ผ3๐‘ฅ๐‘ฅ๏ˆ=3.

Substituting the values of these limits into the equation gives us limsintanlimsinlimtan๏—โ†’๏Šฆ๏Šจ๏Šจ๏Šจ๏—โ†’๏Šฆ๏Šจ๏—โ†’๏Šฆ๏Šจ๏Šจ๏Šจ7๐‘ฅ+33๐‘ฅ8๐‘ฅ=18๏€ฝ๏€ฝ7๐‘ฅ๐‘ฅ๏‰๏‰+38๏€ผ๏€ผ3๐‘ฅ๐‘ฅ๏ˆ๏ˆ=18(7)+38(3)=498+278=768=192.

Therefore, limsintan๏—โ†’๏Šฆ๏Šจ๏Šจ๏Šจ7๐‘ฅ+33๐‘ฅ8๐‘ฅ=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 limsin๏—โ†’๏‘ฝ๏Žก2โˆ’2๐‘ฅ4๐‘ฅโˆ’2๐œ‹.

Answer

Since this is the limit of a trigonometric and algebraic expression, we can attempt to evaluate this limit by direct substitution: 2โˆ’2๏€ป๏‡4๏€ป๏‡โˆ’2๐œ‹=00.sin๏Ž„๏Šจ๏Ž„๏Šจ

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: limsinlimsinlimsin๏—โ†’๏—โ†’๏Ž„๏Šจ๏—โ†’๏Ž„๏Šจ๏‘ฝ๏Žก๏‘ฝ๏Žก๏‘ฝ๏Žก๏€ฝ2โˆ’2๐‘ฅ4๐‘ฅโˆ’2๐œ‹๏‰=๏‚2(1โˆ’๐‘ฅ)4๏€ป๐‘ฅโˆ’๏‡๏Ž=๏‚1โˆ’๐‘ฅ2๏€ป๐‘ฅโˆ’๏‡๏Ž.

To evaluate this limit, we will use the substitution ๐œƒ=๐‘ฅโˆ’๐œ‹2. As ๐‘ฅ approaches ๐œ‹2, ๐œƒ will approaches 0. This gives us limsinlimsin๏—โ†’๏Ž„๏Šจ๏ผโ†’๏Šฆ๏Ž„๏Šจ๏‘ฝ๏Žก๏‚1โˆ’๐‘ฅ2๏€ป๐‘ฅโˆ’๏‡๏Ž=๏‚1โˆ’๏€ป๐œƒ+๏‡2๐œƒ๏Ž.

Recall that sincos๏€ป๐œƒ+๐œ‹2๏‡โ‰ก๐œƒ. We can use this to rewrite the limit as limsinlimcoslimcos๏ผโ†’๏Šฆ๏Ž„๏Šจ๏ผโ†’๏Šฆ๏ผโ†’๏Šฆ๏‚1โˆ’๏€ป๐œƒ+๏‡2๐œƒ๏Ž=๏€ฝ1โˆ’๐œƒ2๐œƒ๏‰=12๏€ฝ1โˆ’๐œƒ๐œƒ๏‰.

Finally, we recall that limcos๏—โ†’๏Šฆ1โˆ’๐‘Ž๐‘ฅ๐‘ฅ=0.

Hence, 12๏€ฝ1โˆ’๐œƒ๐œƒ๏‰=12ร—0=0.limcos๏ผโ†’๏Šฆ

Therefore, limsin๏—โ†’๏‘ฝ๏Žก2โˆ’2๐‘ฅ4๐‘ฅโˆ’2๐œ‹=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 limcotcsc๏—โ†’๏Šฆ๏Šจ6๐‘ฅ4๐‘ฅ8๐‘ฅ.

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: limcotcsclimlimsintan๏—โ†’๏Šฆ๏Šจ๏—โ†’๏Šฆ๏Šง๏Šช๏—๏Šง๏Šฎ๏—๏—โ†’๏Šฆ๏Šจ6๐‘ฅ4๐‘ฅ8๐‘ฅ=๏ƒ6๐‘ฅ๏€ป๏‡๏€ป๏‡๏=๏€ฝ6๐‘ฅ8๐‘ฅ4๐‘ฅ๏‰.tansin๏Žก

We can then rewrite this in terms of the limit results. If ๐‘ฅ is measured in radians and ๐‘Žโˆˆโ„, limsinlimtan๏—โ†’๏Šฆ๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž,๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž.

So, we have limsintanlimsintanlimtanlimsinlimtanlimsinlimtanlimsinlimtanlimsin๏—โ†’๏Šฆ๏Šจ๏—โ†’๏Šฆ๏Šจ๏Šจ๏—โ†’๏Šฆ๏Šจ๏Šจ๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏Šจ๏Šจ๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏Šจ๏—โ†’๏Šฆ๏—โ†’๏Šฆ๏Šฑ๏Šจ๏—โ†’๏Šฆ๏€ฝ6๐‘ฅ8๐‘ฅ4๐‘ฅ๏‰=๏€พ6๐‘ฅ8๐‘ฅ๐‘ฅ4๐‘ฅ๏Š=๏€พ6๐‘ฅ4๐‘ฅ๏Š๏€ฝ8๐‘ฅ๐‘ฅ๏‰=6๏€พ๐‘ฅ4๐‘ฅ๏Š๏€ฝ8๐‘ฅ๐‘ฅ๏‰=6๏€ผ๏€ป๐‘ฅ4๐‘ฅ๏‡๏ˆ๏€ฝ8๐‘ฅ๐‘ฅ๏‰=6๏€ผ๏€ผ4๐‘ฅ๐‘ฅ๏ˆ๏ˆ๏€ฝ8๐‘ฅ๐‘ฅ๏‰.

Applying the limit results, we conclude that 6๏€ผ๏€ผ4๐‘ฅ๐‘ฅ๏ˆ๏ˆ๏€ฝ8๐‘ฅ๐‘ฅ๏‰=6(4)(8)=3.limtanlimsin๏—โ†’๏Šฆ๏Šฑ๏Šจ๏—โ†’๏Šฆ๏Šฑ๏Šจ

Hence, limcotcsc๏—โ†’๏Šฆ๏Šจ6๐‘ฅ4๐‘ฅ8๐‘ฅ=3.

Letโ€™s finish by recapping some of the important points from this explainer.

Key Points

  • We can evaluate the limit of any trigonometric function at ๐‘ฅ=๐‘Ž by direct substitution if a is in its domain.
  • If ๐‘ฅ is measured in radians, we have the following trigonometric limit results:
    • limsin๏—โ†’๏Šฆ๐‘ฅ๐‘ฅ=1,
    • limtan๏—โ†’๏Šฆ๐‘ฅ๐‘ฅ=1,
    • limcos๏—โ†’๏Šฆ1โˆ’๐‘ฅ๐‘ฅ=0.
  • If ๐‘ฅ is measured in radians and ๐‘Žโˆˆโ„, we have the following trigonometric limit results:
    • limsin๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž,
    • limtan๏—โ†’๏Šฆ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž,
    • limcos๏—โ†’๏Šฆ1โˆ’๐‘Ž๐‘ฅ๐‘ฅ=0.
  • We can use these results to evaluate the limit of trigonometric functions.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.