Question

for the function sinc(x)=sinx/x , why is the point x=0 defined ?
I know the limit as x->0 of sinx/x=1 , but still there is a 0 in the denominator at x=0 , so how does this work ?

Answer 1

I think that the answer is that sinc would be discontinuous at x = 0, if sinc wasn't defined to be 1 at 0. Sinc does what you would expect except at zero where it is defined so as to make it useful for making things happen.

Answer 2

I thought that too , since the discontinuity is removable , but then shouldn't sinc(x) be defined in 2 different pieces , namely sinx/x for x different than 0 , and 1 for x=0 ?

Answer 3

Again, I'm guessing here, but sinc is defined piecewisely; 1 is the value when x=0, sinx/x otherwise. Isn't it a matter of what you want the function to do? In some cases, sinc without a value at zero might be useful, in others, not so much?

Answer 4

perhaps , but the way I encountered the function was simply sinc(x)=sinx/x , without the 1 at x=o part , so that's why I am asking : ))

Answer 5

I see your point, indeed.... Are you watching 18.01 at MIT's Open Course Ware?

Answer 6

yes, the value of sinc(0) is defined to be the limit x->0 of sinc(x)
i.e. 1

Answer 7

@against.names yeah , I've followed all the lectures a year ago , but I found out about the 18.01SC just recently , so I thought I'd do the practice problems and the problem sets :))

@phi what about the ctg(x) at x=0 ? I never see it defined separately for x=0

Answer 8

I guess it's because when x->0, sin(x) ->0 at the same time(and when x=0, sin(x)=0 too), that's why sin(x)/x=1