According to the 7th representation in this site:
$$\lim_{M\to \infty} \frac{1}{2\pi\sin(\omega/2)}\sin\left(\omega\left(M+\frac12\right)\right) = \delta(\omega)$$
I'm trying to understand why this is true. I haven't been able to find anything on the Internet.
I must state that I have a very shallow knowledge about distribution theory, so the simpler the explanation, the better. If there is no way to show why that limit is correct without using complicated arguments, go ahead anyway, I'll try to follow up.
The expression you are dealing with is known as the Dirichlet kernel. One defines:
$$D_n(x) = \sum_{k=-n}^{+n} e^{ikx} = 1+\sum_{k=1}^n 2\cos(kx) $$
This expression can be further simplified, using Chebyshev polynomials, or one of the methods in the wikipedia article to obtain
$$D_n(x) = \frac{\sin ((n+\tfrac 12)x)}{\sin(\tfrac 12x)} $$
Now note that $D_n$ by itself 'converges' against the fourier series $\sum_{k=-\infty}^{+\infty} A_k e^{ikx}$, where all the coefficients are $1$. To see that this is related to the dirac delta, take a $2\pi$-periodic function $f$ and compute the convolution $D_n * f$:
$$ D_n *f (x) = \int_{-\pi}^{+\pi} f(y) D_n(x-y)dy = \sum_{k=-n}^{+n}\Big[\int_{-\pi}^{+\pi}f(y)e^{iky} dy\Big]e^{ikx}$$
But on the other hand, the fourier series representation of $f$ is $f(x) = \sum_{k=-\infty}^{+\infty}\hat f(k) e^{ikx}$, where $\hat f(k) = \frac{1}{2\pi} \int_{-\pi}^{+\pi} f(x) e^{ikx}dx$. So we conclude that
$$ \lim_{n\to\infty} D_n * f = 2\pi f$$
For all $2\pi$ periodic function that can be expressed by a fourier series!
But as you may know the dirac delta has the unique property that $\delta * f= f$ for any function $f$, i.e. it acts as the identity element with respect to convolution (this is a direct consequence of $\int f(x)\delta(x)dx = f(0)$).
Thus $\lim_{n\to\infty} \frac{1}{2\pi} D_n(x) = \delta(x)$
Finally one thing that is important to note is that since everything was done in the setting of $2\pi$ periodic functions, the above dirac delta is actually a $2\pi$ periodic version of the 'ordinary' dirac delta. (obvious also by the fact that $D_n$ is $2\pi$ periodic!)