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In the context of working through a proof of the Invariant Set Theorem for continuous-time dynamical systems, I hit a minor technical issue.

Let ${\bf x} : [0,\infty) \rightarrow {\cal R}^d$ be the trajectory of a time-invariant dynamical system whose vector field is a continuous function ${\bf f} : {\cal R}^d \rightarrow {\cal R}^d$. Let $V : {\cal R}^d \rightarrow {\cal R}$ be a continuously differentiable function. Assume that we know that $V({\bf x}(t))$ converges to a constant $V^*$ as $t \rightarrow \infty$. Let $v : {\cal R} \rightarrow {\cal R}$ and define $v(t) = V({\bf x}(t))$. Then note: $dv/dt = [dV({\bf x})/d{\bf x}] {\bf f}({\bf x}(t))$. I would like to show that $dv/dt \rightarrow 0$ as $t \rightarrow \infty$. Intuitively this seems correct and this argument is commonly used but I'm having a mental block regarding how to make it rigorous.

One approach, for example, would be to write: $\lim_{t \rightarrow 0} \frac{dv}{dt} = \lim_{h \rightarrow 0} \lim_{t \rightarrow 0} \frac{ v(t+h) - v(t)}{h} = 0$

But I'm not sure if it is legitimate to interchange the two limit operators? If this is not legitimate? Could someone provide an alternative suggestion? If the limit operator interchange is legitimate, then could someone give me a quick pointer as to how I can verify it is legitimate for myself?

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    As stated this is not true in general. Counterexample: $\dot{x}=1$, $V(x)=\frac{1+sin(x^3)}{1-x(0)+x}$. For this example $\lim_{t\rightarrow\infty}V(x(t))=0$ but $\dot{V}$ does not converge to zero. – 2017-02-24

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As $\dot v(t)=V'(x(t))\,f(x(t))$ combines two continuous functions in $x$ it also has a limit for $t\to\infty$. Now fix some $h$, $h=1$ for instance, and apply the mean value theorem $$ v(t+h)-v(t)=v'(t+\theta_t\,h),\quad \theta_t\in(0,1) $$ which tells you that at least for the sequence $t_n=(n+\theta_{nh})h$ the sequence $v(t_n)$ converges to zero.