Question

In session 8 problem, it says that sinc(0) = 1. Wouldn't it be more correct to say that we define sinc(0) = 1, based on the sinx / x limit? (sinx / x) (0) does not exist so you can't just say that sinc(0) *is* equal to 1, unless you make it to be so as you define the function.

Answer 1

***Wouldn't it be more correct to say that we define sinc(0) = 1,***

It is always nice to give background on why things are defined as they are... but once we have done that, it's ok to just say sinc(0)=1 (side-tracks or tangents to the discussion are not always helpful)

Answer 2

Well, I generally agree with what you say, but this is addressed to beginners, so it wouldn't harm in this case to be a little more specific, in my opinion. It all depends on the context, and for the specific course it could be confusing. And you wouldn't need a side-track either: just a change in the expression to cover any potential misunderstandings.

Answer 3

I agree with atas: Not to be pedantic, but "sloppy" is often equivalent to "confusing". The heuristic proofs [i.e. "hand-waving"] are OK for giving a sense of the proofs of the trig. limits but are ultimately unsatisfying. Saying there is a my analytically precise proof to come in a later course would suffice. We are following an MIT course after all.

For instance in the solution, one might have said that at x =0 we have a removable discontinuity and hence define sinc(0) =1. and notice the solution states that "When x <0 both 1/x and sin(x) are negative..." Is this true on the ENTIRE domain of sinc(x)? Is sin(x) REALLY negative ffor ALL radian values of x<0?