anonymous
  • anonymous
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 ?
OCW Scholar - Single Variable Calculus
schrodinger
  • schrodinger
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anonymous
  • anonymous
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.
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
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?

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anonymous
  • anonymous
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 : ))
anonymous
  • anonymous
I see your point, indeed.... Are you watching 18.01 at MIT's Open Course Ware?
phi
  • phi
yes, the value of sinc(0) is defined to be the limit x->0 of sinc(x) i.e. 1
anonymous
  • anonymous
@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
mytyl
  • mytyl
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

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