Before I give the problem, I want posters to know I am NOT asking for the answer, so don't give it to me. Nor should you give away the entire process, because all I have to then do is plug in numbers, so you are giving me the answer anyway. I just want where to start and to be let known if I am doing the problem incorrectly. So, the problem I am assigned is: "analytically simplify the following limit, which represents the definition of the derivative f'(a) for the function f(x)=sinx and x=a." My teacher doesn't give the limit, and so I don't know where to start. The substitute (since the teacher wasn't there yesterday) said something about about the "angle addition" rule or something, so what I have so far is: (sin(x+dx)-sinx)/(dx)-->(sinxcosdx+sindxcosx-sinx)/(dx)-->(sinx-sinx+sindxcosx)/(dx)-->(sindxcosx)/(dx)-->(sin0cosx)/0=limit does not exist but the limit should definitely exist at all times except when sinx-->0, which it isn't necessarily doing.

The limit as dx approaches 0 exists, and is equal to the derivative of sin(x), which is cos(x). To simplify, you will need to use these two facts: 1.) lim x--> 0 sin(x)/x = 1 2.) sin(x) - sin(y) = 2cos(1/2(x + y))sin(1/2(x-y)) So, lim dx --> 0 (sin(x+dx) - sin(x))/dx = lim dx--> 0 ((2cos(x + dx/2)sin(dx/2))/dx = lim dx --> 0 cos(x+dx/2)sin(dx/2)/(dx/2) = cos(x + 0/2) * 1 = cos(x). Hence, d/dx (sin(x)) = cos(x).

I would have said, "You're on the right track, but your evaluation of the limit of (sin x)/x needs more attention." Instead, Manus now has the answer in its full glory, which he did not want .

This question is extensively discussed on the Web, including the fact that the proof of the limit of sin(x)/x as x->0 is 1 should come from some method (like geometric analysis) rather than L'Hospital's rule, otherwise it will become a circular proof.