Delta function equivalent : limit $\sin(t/e)/(\pi t)$ tends to delta(t) as $e$ approaches $0$

Question

I plotted the above function in MATLAB and observed the delta function, but I can't theoretically prove that limit of $$\frac{\sin(t/\varepsilon)}{\pi t}$$ tends to delta when $\varepsilon$ approaches $0$. As we choose $t$ near $0$ we get infinity as the result, but I can't prove where ever except $t=0$ we will have zero output.

Answer 1

A sequence $(\delta_\alpha)_{\alpha\in\mathbb{N}}$ of integrable functions $\delta_\alpha\in L^1(\mathbb{R}^n)$ is called a nascent delta function if

1. for all $x\in\mathbb{R}$ and $\alpha\in\mathbb{N}$, $(\delta_\alpha)_{\alpha\in\mathbb{N}}\ge0,$
2. for all $\alpha\in\mathbb{N}$, $$\int_{\mathbb{R}^n}\delta_{\alpha}(x)\,\mathrm{d}x=1,$$
3. for all $\varepsilon$, $$\lim_{\alpha \to \infty} \int_{\mathbb{R}^n \setminus B_\varepsilon (0)} \delta_{\alpha}(x) \mathrm{d}x = 0.$$

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The function $$\delta_{\epsilon}(x)=\frac{1}{\pi x}\sin\left(\frac{x}{\epsilon}\right)$$ is not a nascent Delta-function since it is not positive for all $\varepsilon\ge0$. Still the other two conditions are met so it tends to the Delta-function.

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