Limit of $ \frac{1}{2}\frac{\sin^2(\omega t/2)}{(\omega/2)^2}$

Question

I'm trying to prove that

$\lim_{t\rightarrow \infty} \frac{1}{2}\frac{\sin^2(\omega t/2)}{(\omega/2)^2} = \pi \delta(\omega) t $

Any thoughts on how could I do that? I thought about mapping it to the identity $\int_{-\infty}^{\infty}dx\frac{\sin\left(x\right)}{x}=\pi$ but I did not succeed.

Thanks.