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## AP®︎ Calculus AB (2017 edition)

### Unit 1: Lesson 9

Trigonometric limits & squeeze theorem- Limits of trigonometric functions
- Limits of trigonometric functions
- Trig limit using Pythagorean identity
- Trig limit using double angle identity
- Limits using trig identities
- Squeeze theorem intro
- Squeeze theorem example
- Squeeze theorem
- Limit of sin(x)/x as x approaches 0

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# Limits of trigonometric functions

AP.CALC:

LIM‑1 (EU)

, LIM‑1.D (LO)

, LIM‑1.D.1 (EK)

Like other common functions, we can use direct substitution to find limits of trigonometric functions, as long as the functions are defined at the limit.

## Want to join the conversation?

- Is Sal using Degrees, Radians or Gradians?(25 votes)
- Following Nathaniel's answer, note that the widely taught slopes of graphs of trigonometric functions only work in radians. In fact, many facts involving derivatives of trigonometric functions only hold if angles are measured in radians.

It is helpful to remember that radians are the more natural way to measure angles when compared to degrees; humans chose 360 degrees for a complete rotation because 360 is close to 365, the number of days in a year, or simply because 360 has a lot of divisors.

Finally, the radian is a dimensionless quantity, a pure number.(22 votes)

- In the "Limits of trigonometric functions" video, it is stated that all trig functions are continuous. How is tan x continuous when there are asymptotes?(18 votes)
- I'm going to attempt to answer this, even though I suspect you may feel cheated by my answer.

What Sal actually says in the video is:0:34in fact, all of the trigonometric functions are continuous0:37**over their entire domain**.

And later says:3:58And one way to think about it is pi over two4:01is**not in the domain of tangent of x**.

Which is a bit like saying they're continuous except where they're not. I admit, I feel a little bit cheated too, but in the context of being able to find limits it makes sense.(44 votes)

- I am having great difficulty with this topic, limits of trigonometric functions. Is there a chart one could refer me to regarding the values of these functions in relation to other trig functions, such as tan = sin/cos?

It seems I could use a refresher on trigonometry as a whole as well.

Any resources you guys could point me to would be greatly appreciated. Thank you.(11 votes)- I've attached a link to the beginning of the trigonometry series, below. I'd suggest you start there, and work your way through them all. If you are struggling with trig, it could make some of your work in calculus pretty unpleasant. Good luck, best wishes!

https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/intro-to-the-trig-ratios/a/opposite-adjacent-hypotenuse(11 votes)

- At0:35, aren't all functions, trigonometric or not, continuous over their entire domain? Isn't this built into the definition of a function's domain?

Or is this important because rational functions, defined for all real numbers except asymptotes, can have holes (removable discontinuities), and trigonometric functions never do?(5 votes)- Hi there... This is tricky - because MOST functions we are familiar with ARE continuous over their domain... but there are some funky functions that are not.

One in particular, is a function called the Greatest Integer Function - this function defines for every real number in its domain (that means, the domain is all real numbers) the greatest integer that is less than or equal to it. Let's call this function GIF(x).

GIF(0) = 0, GIF(1) = 1, GIF(2) = 2, etc... input an integer, output that integer.

GIF(0.1) --> think of all integers less than 0.1 - which is the greatest of those? 0.

so, GIF(0.1) = 0, GIF(0.2) = 0, GIF(0.9) = 0, GIF(0.999999) = 0

BUT

GIF(1) = 1 and by the same definition, GIF(1.1) = 1, GIF(1.2) = 1, etc

So by the definition of continuity at a point, the left and right hand limits of the GIF function at integers will always be different - therefore, no limit will exist at the integers, even though integers are in the domain of the function.

Hope this helps :)(11 votes)

- Why did you take the value of cos pi as -1? My calculator says it's +0.9985(7 votes)
- He was referring to the value pi in radians which converts to 180 in degrees. Your calculator gave you the value of cos of pi degrees, you need to enter cosine of pi
**radians**in your calculator(8 votes)

- Is a trigonometric function defined over complex numbers? For example, is it possible to have sin(sqrt{-1})?(3 votes)
- Great question!

In ordinary trigonometry, the answer is no.

However, if we extend Euler's formula e^(iz)=cos(z) + i sin(z) to complex-valued z, then the answer is yes!

We have

e^(i*i) = cos(i) + i sin(i) and

e^(i*-i) = cos(-i) + i sin(-i).

Recall that cosine and sine are even and odd functions, in this order. If we extend this concept to complex numbers and also use the fact that i^2=-1, we can rewrite the pair of equations as

e^(-1) = cos(i) + i sin(i) and

e^(1) = cos(i) - i sin(i).

Subtracting the second equation from the first equation and then dividing by 2i, we have

sin(i) = [e^(-1)-e^(1)]/(2i) = [-(1/e - e)/2]*i = [(e - 1/e)/2]*i.

So sin(sqrt{-1}) = [(e - 1/e)/2]*i.(3 votes)

- how do yo put co secant and cotangent on a calculator(3 votes)
- For cosecant, you enter 1/sine. For cotangent, you enter 1/tangent. For the other reciprocal trigonometric function, secant, you enter 1/cosine.(1 vote)

- How would you change lim x->2 1/x-2 so that the problem is not 0/0?(3 votes)
- Your limit is
**not**indeterminate (`0/0`

) – the numerator is 1!(3 votes)

- at2:39, why didn't Sal just substitute x = pi and solve for the limit of tan(pi)?(2 votes)
- Because he had not yet established that tan(x) is continuous at π, so he needed to deduce that from the fact that sine and cosine are continuous.(3 votes)

- how do I solve this [(x-1)^2 cos(1/(x-1)^2)]. lim x->1(2 votes)
- lim [(x-1)^2 cos(1/(x-1)^2)]

lim (x-1)^2 * lim (cos(1/(x-1)^2))

0 * lim (cos(1/(x-1)^2))

Then, since we just have the second part to deal with we can use the squeeze theorem. We know cos(x) will always be bounded by -1 and 1 no matter how large or small the value within is, so we know we can just multiply it by 0 to get the whole limit goes to 0

Finding the actual limit of lim (cos(1/(x-1)^2)) is not something you can do, but since you know that the numbers exist everywhere around 1 they can be multiplied by 0. I hope that made sense.(1 vote)

## Video transcript

- [Instructor] What we're
going to do in this video is think about limits involving
trigonometric functions. So let's just start with a
fairly straightforward one. Let's find the limit as x
approaches pi of sine of x. Pause the video and see if
you can figure this out. Well, with both sine of x and cosine of x, they are defined for all real numbers, so their domain is all real numbers. You can put any real number in here for x and it will give you an output. It is defined. And they are also continuous
over their entire domain, in fact, all of the trigonometric
functions are continuous over their entire domain. And so for sine of x,
because it's continuous, and is defined at sine of pi, we would say that this
is the same thing as sine of pi, and sine of pi, you might already know, is equal to zero. Now we could do a similar
exercise with cosine of x, so if I were to say what's
the limit as x approaches, I'll just take an arbitrary
angle, x approaches pi over four of cosine of x? Well once again, cosine of x is defined
for all real numbers, x can be any real number. It's also continuous. So for cosine of x, this
limit is just gonna be cosine of pi over four, and that is going to be equal to square root of two over two. This is one of those useful angles to know the sine and cosine of. If you're thinking degrees,
this is a 45 degree angle. And in general, if I'm dealing
with a sine or a cosine, the limit as x approaches a of sine of x is equal to sine of a. Once again, this is going
to be true for any a, any real number a. And I can make a similar
statement about cosine of x. Limit as x approaches a of cosine of x is equal to cosine of a. Now, I've been saying it over and over, that's because both of their
domains are all real numbers, they are defined for all
real numbers that you put in, and they're continuous
on their entire domain. But now, let's do slightly more involved trigonometric functions,
or ones that aren't defined for all real numbers, that
their domains are constrained just a little bit more. So let's say if we were to take the limit as x approaches pi of tangent of x. What is this going to be equal to? Well, this is the same thing as the limit as x approaches pi. Tangent of x is sine
of x over cosine of x. And so both of these are defined for pi and so we could just substitute pi in. And we just wanna ensure
that we don't get a zero in the denominator, because
that would make it undefined. So we get sine of pi over cosine of pi which is
equal to zero over negative one, which is completely fine. If it was negative one over
zero, we'd be in trouble. But this is just gonna be equal to zero. So that works out. But if I were to ask
you, what is the limit as x approaches pi over
two of tangent of x? Pause the video and try to work that out. Well, think about it. This is the limit as x
approaches pi over two of sine of x over cosine of x. Now sine of pi over two is one, but cosine of pi over two is zero. So if you were to just substitute in, this would give you one over zero. And one way to think
about it is pi over two is not in the domain of tangent of x. And so this limit actually
turns out, it doesn't exist. In general, if we're
dealing with the sine, cosine, tangent, or cosecant,
secant, or cotangent, if we're taking a limit to a point that's in their domain,
then the value of the limit is going to be the same thing as the value of the function at that point. If you're taking a limit to a point that's not in their domain, there's a good chance that
we're not going to have a limit. So here, there is no limit. And the way to do that is that pi over two is not in tangent of x's domain. If you were to graph
tan of x, you would see a vertical asymptote at pi over two. Let's do one more of these. So let's say the limit as x
approaches pi of cotangent of x, pause the video and see
if you can figure out what that's going to be. Well, one way to think about it, cotangent of x is one over tangent of x, it's cosine of x over sine of x. This is a limit as x approaches pi of this. And is pi in the domain of cotangent of x? Well, no, if you were
just to substitute pi in, you're gonna get negative one over zero. And so that is not in the
domain of cotangent of x. If you were to plot it, you
would see a vertical asymptote right over there. And so we have no limit. We have no limit. So once again, this is
not in the domain of that, and so good chance that we have no limit. When the thing we're taking the limit to is in the domain of the
trigonometric function, we're going to have a defined limit. And sine and cosine in
particular are defined for all real numbers
and they're continuous over all real numbers. So you take the limit
to anything for them, it's going to be defined
and it's going to be the value of the function at that point.