In lecture 3, when Professor David was proving that:
limit (as Delta x goes to 0) of sin Delta x over Delta x = 1
He said that x here must be measured in radians, so why it must be measured in radians not degrees ?

Question

In lecture 3, when Professor David was proving that:
limit (as Delta x goes to 0) of sin Delta x over Delta x = 1
He said that x here must be measured in radians, so why it must be measured in radians not degrees ?

phi · Answer

we can give a geometric proof of this limit. see https://www.khanacademy.org/math/differential-calculus/limits_topic/squeeze_theorem/v/proof-lim-sin-x-x notice that we compare the length of an arc (along the circle) to straight lines. the length of an arc is $$ s = r \theta$$ if theta is measured in radians. if we use a unit circle, with radius r =1, then the length along the arc becomes $$ s = \theta$$ if we measure theta in degrees, this is not true, wiithout some correct factor.

anonymous · Answer

By the last line in your reply, do you mean that this proof also holds if we used degrees from the beginning ?

phi · Answer

if you use degrees, you would need a "correction factor"  
it would not be  sin(x)/x  but   sin(x)/kx  where k corrects (converts)  x from degrees to radians.  Of course, this is not very convenient.