Why does the sinc function approach zero as the argument approaches infinity?

[McLaren Rulez]

Consider sin(vx)/x as v 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 sin(vx)/x 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.

 

[GarageDweller]

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.

 

[McLaren Rulez]

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.

 

[Mute]

The proper way to show that

\lim_{v \rightarrow \infty} \frac{\sin(vx)}{x} = \delta(x)

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

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

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 "\lim_{v \rightarrow \infty} \frac{\sin(vx)}{x} = \delta(x)".

 

[McLaren Rulez]

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.