In proofs of Lyapunov's direct method (see this lecture, pg 8 for example), the claim is made that because of continuity of $V$, to show $x(t) \rightarrow 0$, it suffices to show that $V(x(t)) \rightarrow 0$. Why is this true?
Continuity implies that if $\Vert x \Vert < \delta$ then $V(x) <\epsilon$ for appropriate $(\delta, \epsilon)$, but it seems like here we want to show that another relation also holds, that if $V(x) < \delta_0$ then $\Vert x \Vert < \epsilon_0$ for some appropriate $(\delta_0, \epsilon_0)$.
I'm aware that if $g(t)$ converges as $t \rightarrow \infty$ and $f$ is continuous then $\lim_{t \rightarrow \infty} f(g(t)) = f(\lim_{t \rightarrow \infty} g(t))$ but I don't think we can assume $x(t)$ converges: that's one of the results we are trying to prove!