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Limits of trigonometric functions

This video covers limits of trigonometric functions, focusing on sine, cosine, and tangent. It emphasizes that sine and cosine are continuous and defined for all real numbers, so their limits can be found using direct substitution. For tangent and cotangent, limits depend on whether the point is in their domain.

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  • piceratops ultimate style avatar for user Elijah Gregg
    Is Sal using Degrees, Radians or Gradians?
    (31 votes)
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    • aqualine seed style avatar for user William Park
      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.
      (30 votes)
  • blobby green style avatar for user Mary Beth Cucinotta
    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?
    (20 votes)
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    • leaf yellow style avatar for user Howard Bradley
      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:
      in fact, all of the trigonometric functions are continuous
      over their entire domain.

      And later says:
      And one way to think about it is pi over two
      is 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.
      (52 votes)
  • purple pi purple style avatar for user Keith
    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.
    (14 votes)
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  • duskpin tree style avatar for user G Y
    At , 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)
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    • duskpin ultimate style avatar for user Ms Demaray
      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 :)
      (14 votes)
  • starky ultimate style avatar for user Isuru Udayanga Rathnayake
    Why did you take the value of cos pi as -1? My calculator says it's +0.9985
    (7 votes)
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  • blobby green style avatar for user Elsa
    This is probably a dumb question, but how on earth is he getting the square root of 2 over 2? When I type in the cos(pi/4) I get 0.7071... The basic concept of this makes sense, but I'm not getting fraction numbers.
    (4 votes)
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    • male robot donald style avatar for user Venkata
      Take a triangle with the height and base equal to 1. Now, using the Pythagorean Theorem, you can find out that the hypotenuse = root(2). Now, take the cosine of any of the 45 degree angles. You'll get cos(x) = adjacent/hypotenuse, which gives 1/root(2) or, on rationalizing, root(2)/2.

      This can also be derived from the unit circle, which I recommend you check out if you're unfamiliar with how trig ratios are derived.

      Also, never hesitate to ask questions. There's no such thing as a "dumb" question when you're learning! :)
      (9 votes)
  • aqualine ultimate style avatar for user 103429
    As soon as I saw the first practice problem in the video, I cowered.
    (8 votes)
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  • duskpin ultimate style avatar for user Daniel Marques
    Is a trigonometric function defined over complex numbers? For example, is it possible to have sin(sqrt{-1})?
    (3 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      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.
      (5 votes)
  • blobby green style avatar for user Daeelhawk
    how do yo put co secant and cotangent on a calculator
    (3 votes)
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  • blobby green style avatar for user Lyle Batley
    Sal needs to state that this is in radians.
    (4 votes)
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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.