# Mathematical proof and intuitive reasoning for: $\lim \limits_{\epsilon \to 0} \frac{1}{\pi t}\sin\frac{\pi t}{\epsilon}=\delta(t)$

Please provide the mathematical proof and intuitive reasoning for:

$$\lim \limits_{\epsilon \to 0} \frac{1}{\pi t}\sin\frac{\pi t}{\epsilon}=\delta(t)$$ Where $\delta(t)=$Unit impulse function

Intuitive reasoning (what I have gathered):

If we compare $\sin\frac{\pi t}{\epsilon}$ with $\sin(\omega t)$, then we can see that with decreasing value of $\epsilon$ the waveform $\sin\frac{\pi t}{\epsilon}$ approaches a sine wave of very high frequency.

Along with that, $\frac{1}{\pi t}$ as a multiplicative factor implies that at $t=0$, the function would be infinity and as the value of $t$ increases, the amplitude of $\sin\frac{\pi t}{\epsilon}$ goes on decreasing with the function tending to zero as t tends to infinity.

Is my reasoning correct?

• Ok you are right, you want to show that a scaled and stretched sinc pulse ( $\text{sinc}(x) = \frac{ \sin(\pi x) }{\pi x}$) would go to an impulse in the limiting case: $$\lim_{\epsilon \to 0} \frac{1}{\epsilon} \text{sinc}( \frac{t}{\epsilon}) =\lim_{\epsilon \to 0} \frac{ \sin( \pi \frac{t}{\epsilon}) } { \pi t}$$ would intuitively go to an impulse $\delta(t)$. – Fat32 Oct 18 '17 at 16:07
• @Jazzmaniac Sir, I understood your reasoning that how the width and the height of the rectangular pulse is $a$ and $\frac{1}{a}$ respectively and how the limiting condition narrows the width of the pulse and increases its height. I also understood your mathematical proof. The only two things I didn't understand was why for a continuous function $f(x)=f(0)+x \cdot f'(0) + \mathcal{O}(x^2)$ holds. And Edit 1, wherein you have mentioned how to verify the identity in question for values of x not equal to zero. Apart from these two, I felt as if I was gliding on butter! Thank you sir, – Soumee Oct 18 '17 at 21:56