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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?

• 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.
• 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?
• 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.
• 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.
• 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?
• 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 :)
• Why did you take the value of cos pi as -1? My calculator says it's +0.9985
• 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
• Is a trigonometric function defined over complex numbers? For example, is it possible to have sin(sqrt{-1})?
• 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.
• how do yo put co secant and cotangent on a calculator
• 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?
• Your limit is not indeterminate (`0/0`) – the numerator is 1!
• at , why didn't Sal just substitute x = pi and solve for the limit of tan(pi)?
• 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.
• how do I solve this [(x-1)^2 cos(1/(x-1)^2)]. lim x->1