Consider the system $\dot x= f(x)+u$, where $u=e^{-t}$ is a vanishing input.
Let first $f=-x$:
Then the Lyapunov candidate $V=1/2 x^2$ has the following time derivative:
$\dot V= -x^2+xu\leq -x^2+|ux|<0 $ for $|x| >|u|.$ As for increasing $t$, $u$ goes to $0$, also $x$ tends to $0$.
Suppose now that $f=0$:
Repeating the same steps again, clearly, $\dot V\leq |u||x|$.
Is it correct to say that the solution $x(t)$ tends to a constant value (as $\dot V \to 0$), i.e., is bounded as $t$ to $\infty$?
In this example, by setting $x=\sqrt{2 V}$, the differential inequality can be solved which shows that $V$ is bounded as $t \to \infty$.
Moreover, is there a general theorem about the topic of vanishing perturbations and bounded solutions/stable (not asymptotically stable) systems considering Lyapunov methods?
Normally, to say that $x$ is bounded (tends to a constant) you need to show that $V$ is bounded. And saying that $\dot{V}\to0$ is not sufficient, e.g., $\log(t)$.
The properties that you need are the Input-to-State-Stability (ISS), integral Input-to-State-Stability (iISS), and $\mathcal{L}_2$ gain.