Show that every solution $x(t)$ of $x'(t)= A(t)x(t)$ converges to some limit (Long-time asymptotics). Suppose

$$\int_0^∞\|A(t)\|\,dt &lt; ∞.$$

Show that every solution $x(t)$ of $x'(t)= A(t)x(t)$ converges to some limit:
$\lim_{t→∞} x(t) = x_∞.$

(Hint: First show that all solutions are bounded and then use the corresponding integral equation.)

I was able to solve a first part: 

I used that the solutions are limited by $\|\phi(t,t_0)\| \leq e^ {\int_0^∞\|A(t)\|\,dt}$. As $\int_0^∞\|A(t)\|\,dt &lt; ∞$, then $\|\phi(t,t_0)\| \leq K $. The problem now is to show that the above limit always exists. Someone can help
Hint

Once you know that $x$ is bounded by some constant $K$ you can use the integral equation to show that $x(t)$ is Cauchy, for $t_1 \leq t_2$ you have:
$$\|x(t_1)-x(t_2)\| =\left\|\int_{t_1}^{t_2} A(s) x(s) ds \right\| \leq K \int_{t_1} ^{t_2} \|A(s)\| ds$$
And $\int_{t_1} ^{t_2} \|A(s)\|ds$ goes to $0$ as $t_1$ goes to $+\infty$.

So using the sequential characterization of convergence and Cauchy sequences you can show that, as for Cauchy sequences, as long as your space is complete the function converges.
We have $$\frac{x'(s)}{x(s)} =A(s)$$ thus 
$$\ln|x(t)|=\ln |x(0)|+\int_{0}^t\frac{x'(s)}{x(s)}ds =\ln |x(0)| +\int_{0}^t A(s)ds$$ therefore
$$x(t) =\pm x(0) e^{\int_{0}^t A(s)ds}$$ hence
$$\lim_{t\to \infty}x(t) =\lim_{t\to \infty}\left(\pm x(0) e^{\int_{0}^t A(s)ds}\right)=\pm x(0) e^{\int_{0}^{\infty} A(s)ds}$$