Limit of a state-transition matrix Let $\omega$ be a positive real number and let $A,B\in\mathbb{R}^{n\times n}$ be $n\times n$ real matrices. Consider the following linear time-varying dynamical system
$$
\dot{x}(t) = (A + \cos(\omega t)B)x(t),\quad x(0)=x_0\in\mathbb{R}^{n}.
$$
Let $\Phi(t,0)$ denote the state-transition matrix of the above system.

My question. Is it true that
  $$
\lim_{\omega \to \infty} \Phi(t,0) = e^{At} \ \ \ ?
$$

Some remarks. If matrices $A$ and $B$ commute, this is true. Indeed, in this case it holds
$$
\Phi(t,0) = e^{\int_0^t A + \cos(\omega \tau)B\, \mathrm{d}\tau}.
$$
In the non-commuting case, I didn't manage to prove this. The main issue here is that a closed-form expression of $\Phi(t,0)$ does not exist, apparently. The only (perhaps useful) idea that I had so far is to exploit the Peano-Baker expansion of $\Phi(t,0)$. However, even with this tool, I couldn't provide an answer to my question. 
Thanks for your help!
 A: Let me turn my comment into an answer: if we write $x(t)=e^{At}y(t)$, then $y$ satisfies the differential equation $\dot{y}=\cos(\omega t)C(t)y$, where $C(t)=e^{-At}Be^{At}$. We see thus that $\Phi(t,0)=e^{At}\Lambda(t,0)$, where $\Lambda(t,0)$ is the state-transition matrix associated to $\cos(\omega t)C(t)$. Hence, we must prove that $\Lambda(t,0)y_0\to y_0$ when $\omega\to\infty$.
To show this, we integrate the differential equation to get $y(t)=y_0+\int_0^t\cos(\omega s)C(s)y(s)\,ds$, so we must prove that the integral term on the right goes to zero, but $y$ depends on $\omega$, then we write also $y_\omega$. We integrate by parts to get
$$
\frac{1}{\omega}\big[-\int_0^t\sin(\omega s)(C(s)y_\omega(s))'\,ds+\sin(\omega t)C(t)y_\omega(t)\big] \\
= \frac{1}{\omega}\big[-\int_0^t\sin(\omega s)(C'(s)y_\omega(s)+\cos(\omega s)C(s)^2y_\omega(s))\,ds+\sin(\omega t)C(t)y_\omega(t)\big]\hspace{0.5cm}(*)
$$
We want to make $\omega$ goes to infinity, but we must first get a uniform upper bound of $y_\omega$. Take inner product at both sides of the differential equation to reach $\frac{1}{2}\dot{(|y|^2)}=\cos(\omega t)\langle C(t)y,y\rangle\le |C(t)||y|^2$. By Gronwall's inequality we get $|y(t)|\le R(t)|y_0|$, where $R$ is a function independent from $\omega$. Use this in equation (*), to conclude that $y(t)\to y_0$ as $\omega\to\infty$ for fixed $t$, so $\Lambda(t,0)\to I$ as $\omega\to \infty$.
