I am not very sure where you are getting at and what you mean by "frenquency going to $0$". At least in your example the frenquency $\omega_0$ is always a constant.
As I understand, the integral
$$\frac{1}{a-b}\int_a^bf(x)dx$$
is the average of $f(x)$ over the interval $[a,b]$.
If $f(x)$ has period $T$, then one can prove that
$$\frac{1}{T}\int_a^{a+T}f(x)dx\quad(*)$$
is a constant regardless of the choice of $a$. One might say in this case that $f(x)$ has a "well-defined" average.
If, unfortunately, $f(x)$ is not periodic but we still want to talk about the "average" of $f(x)$, we consider its average over some finite interval $[-T/2,T/2]$
$$\frac{1}{T}\int_{-T/2}^{T/2}f(x)dx$$
and take the limit $T\to\infty$ and call the limit (if it exists)
$$\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}f(x)dx$$
the average of $f(x)$ over $\mathbb{R}$.
It can also be shown that the second average is consistent with the first in the case where $f(x)$ is periodic. My opinion is that the limit definition of the average of a function is simply a generalization to deal with not strictly periodic functions. In your case, power is the average of energy over time, so you may understand the limit as the average of the energy of a signal over a very long time.
But still, I don't see how this relates to any change in frenquency.
First thing to note is that ${\rm rect}(t)^2$, at least for the purposes of integration, is equal to ${\rm rect}(t)$. This is trivially true for all $|t|\neq1/2$, and the value at the single points $t=\pm1/2$ has no effect on the value of the integral, as long as no dirac functions are involved. So our integral becomes:
$$P=\frac{1}{T_0}\int_{T_0}(2{\rm rect}(\sin(2\pi t)))^2\ {\rm d}t=
\frac{1}{T_0}\int_{T_0}4{\rm rect}(\sin(2\pi t))\ {\rm d}t$$
Now, the way I'm familiar with handling ${\rm rect}$ and unit step functions in integrals, is to change the bounds. But first, we need to find which values of $t$ over one period satisfy $|\sin(2\pi t)|<1/2$. First note that for one period of the function, it we get two intervals of $t$ that satisfy the inequality (one with $\sin(2\pi t)$ increasing, one decreasing). The endpoints of these intervals can be found by solving the equality:
$$\sin(2\pi t)=\frac{1}{2}\quad\Rightarrow\quad t=\frac{\sin^{-1}(1/2)}{2\pi}=\frac{\pi/6}{2\pi}=\frac{1}{12}$$
$$\sin(2\pi t)=-\frac{1}{2}\quad\Rightarrow\quad t=\frac{\sin^{-1}(-1/2)}{2\pi}=\frac{-\pi/6}{2\pi}=-\frac{1}{12}$$
Remembering that $y=\pi-\sin^{-1}(x)$ is also a solution to $\sin(y)=x$, we get the endpoints of the other interval:
$$\sin(2\pi t)=\frac{1}{2}\quad\Rightarrow\quad t=\frac{\pi-\sin^{-1}(1/2)}{2\pi}=\frac{5\pi/6}{2\pi}=\frac{5}{12}$$
$$\sin(2\pi t)=-\frac{1}{2}\quad\Rightarrow\quad t=\frac{\pi-\sin^{-1}(-1/2)}{2\pi}=\frac{7\pi/6}{2\pi}=\frac{7}{12}$$
Plugging in $T_0=1$ and choosing our period to be $[-1/4,3/4)$, we get:
$$P=
\frac{1}{T_0}\int_{T_0}4{\rm rect}(\sin(2\pi t))\ {\rm d}t=\frac{1}{1}\left(\int_{-1/12}^{1/12}4\ {\rm d}t+\int_{5/12}^{7/12}4\ {\rm d}t\right)\\
=\left(\frac{1}{12}-\frac{-1}{12}\right)4+\left(\frac{7}{12}-\frac{5}{12}\right)4\\
=\left(\frac{2}{12}+\frac{2}{12}\right)4\\
=\frac{4}{3}$$
Best Answer
Hint
For periodic signals $x(t)$ such that $x(t+T_0)=x(t)$ we have $$P_{x(t)}=\lim_{T\to \infty}{1\over 2T}\int_{-T}^T |x(t)|^2dt=\lim_{k\to \infty}{1\over 2kT_0}\int_{-kT_0}^{kT_0} |x(t)|^2dt={1\over T_0}\int_0^{T_0}|x(t)|^2dt$$