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Sinc function limit question

  1. Aug 12, 2012 #1
    Consider [itex]sin(vx)/x[/itex] as [itex]v[/itex] approaches infinity. Now, this becomes a delta function and I have seen graphs that show this function as v increases.

    My question is the following: I cannot quite see why the function [itex]sin(vx)/x[/itex] becomes zero if x≠0. sin(vx) is bounded and oscillates rapidly between -1 and 1 as x is changed. But the envelope of the function is still 1/x so how come it goes to zero for all x≠0? Can someone prove this result?

    Thank you.
  2. jcsd
  3. Aug 12, 2012 #2
    I think you're talking about sin(1/x) not, (sinx)/x

    Edit: Actually re-reading your post, I realize I have no idea what you're asking.

    sinx/x is in no way related to the delta function, and does not oscillate between anything, it's just a normal sinx that diminishes as you approach infinity.
    Sin(1/x) however does oscillate in a divergent manner as you approach 0, the limit is quite strange and nothing is quite like it.
  4. Aug 12, 2012 #3
    No, it is sin(vx)/x. As v gets larger, it approximates a delta function better. Now, with v→∞, I cannot see why the function is zero if x is non zero.

    A similar definition is here http://en.wikipedia.org/wiki/Sinc_function where there is a section on how it relates to the dirac delta function.
    Last edited: Aug 12, 2012
  5. Aug 12, 2012 #4


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    Homework Helper

    The proper way to show that

    [tex]\lim_{v \rightarrow \infty} \frac{\sin(vx)}{x} = \delta(x)[/tex]

    is not to try and show that the left hand side is infinite when x = 0 and zero for non-zero x. That might be true of some representations of the delta function, but really what you need to show is that

    [tex]\lim_{v\rightarrow \infty} \int_{-\infty}^\infty dx~\frac{\sin(vx)}{x} f(x) = f(0).[/tex]

    Similarly you can show that if the integration region does not contain the origin, the integral is zero.

    That is the appropriate sense in which one should interpret "[itex]\lim_{v \rightarrow \infty} \frac{\sin(vx)}{x} = \delta(x)[/itex]".
  6. Aug 13, 2012 #5
    Thank you Mute. Didn't know that before but yeah, I can prove it according to your definition so I guess it's all good.
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