Dirac function squared?

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I am trying to perform the following limit

\begin{equation} \lim_{t\rightarrow\infty}\int\frac{d\omega}{2\pi}S\left(\omega\right)\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}} \end{equation} Ideally, I was thinking about using the relation $\lim_{t\rightarrow\infty}\frac{\sin\left(\pi xt\right)}{\pi x}=\delta\left(x\right)$ to simplify the equation above, and actually this kind of works, but leads to the following integral of a Dirac delta function squared, \begin{align*} \int\frac{d\omega}{2\pi}S\left(\omega\right)\lim_{t\rightarrow\infty}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}} & =\int\frac{d\omega}{2\pi}S\left(\omega\right)\lim_{t\rightarrow\infty}\frac{\sin\left[\pi\left(\frac{\omega+\Omega}{2\pi}\right)t\right]}{\pi\left(\frac{\omega+\Omega}{2\pi}\right)}\frac{\sin\left[\pi\left(\frac{\omega+\Omega}{2\pi}\right)t\right]}{\pi\left(\frac{\omega+\Omega}{2\pi}\right)}\\ & =\int\frac{d\omega}{2\pi}\delta\left[\left(\frac{\omega+\Omega}{2\pi}\right)\right]\delta\left[\left(\frac{\omega+\Omega}{2\pi}\right)\right]\\ & =\int\frac{d\omega}{2\pi}2\pi\delta\left(\omega+\Omega\right)\times2\pi\delta\left(\omega+\Omega\right)\rightarrow\infty \end{align*} I was reading about the square of the Dirac delta function, it turns out this is not even a distribution. [See: https://math.stackexchange.com/questions/2221429/why-is-the-square-of-dirac-delta-function-not-a-distribution];

or even if treated as would diverge after the integration [See: https://physics.stackexchange.com/questions/47934/dont-understand-the-integral-over-the-square-of-the-dirac-delta-function] In a related old post, one of our peers mentioned that is possible to prove

\begin{equation} \lim_{t\rightarrow\infty}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}}=2\pi\delta\left(\omega+\Omega\right)t \end{equation} Limit of $ \frac{1}{2}\frac{\sin^2(\omega t/2)}{(\omega/2)^2}$which resolves easly my problem. However, I do not know how to prove that. It seems one only applies the relation above once.