\section{Lyapunov Like Analysis}
\begin{lemma}
Given a differentiable function $f=f(t):\mb{\Bb R} \to \mb{\Bb R}$ such that $\dot{f}$ is uniformly continuous. If $f$ has a finite limit as $t \to \infty$, then $\dot{f}(t) \to 0$ as $t \to \infty$.
\end{lemma}
{\bf Proof}\quad We prove the lemma by contraposition. We assume that $\dot{f}(t)$ is bounded away from $0$ as $t\to \infty$. Then, there is an $\varepsilon >0$ so that for any $T>0$ there exists $t>T$ such that $|\dot{f}(t)|\ge \varepsilon$. Therefore, we can construct an infinite sequence of $t_i$'s such that $t_i \to \infty$ as $i\to \infty$, and for each $t_i$ we have $|\dot{f}(t_i)|\ge \varepsilon$. By assumption $\dot{f}$ is uniformly continuous. Hence, given $\varepsilon/2>0$ there is $\eta>0$ so that for any $t^{\prime}$ and $t^{\prime \prime}$ such that $\left|t^{\prime} - t^{\prime \prime}\right| < \eta$ we have
\[
\left|\: \left|\dot{f}(t^{\prime} )\right|-\left|\dot{f}(t^{\prime \prime} )\right|\:\right|\le\left| \dot{f}(t^{\prime} )-\dot{f}(t^{\prime \prime} )\right| < \frac{\varepsilon}{2}.
\]
The above implies that the value of $\dot{f}(t)$ over an interval $[t_i-\eta,t_i+\eta]$ cannot change by more than $\varepsilon/2$. Hence, for any $t$ such that $|t-t_i|<\eta$, we have
\[
\left| \dot{f}(t)\right| > \frac{\varepsilon}{2}.
\]
Using the above, we obtain
\begin{eqnarray*}
\left| \int_{t_i-\eta}^{t_i+\eta}\dot{f}(t)dt \right|&=& \int_{t_i-\eta}^{t_i+\eta}\left|\dot{f}(t)\right|dt\\
&\ge& \frac{\varepsilon}{2}2\eta\\
&=& \varepsilon \eta.
\end{eqnarray*}
However, the above contradicts the fact that
\[
\int_0^{\infty}\dot{f}(t)dt=f(\infty)-f(0)<\infty,
\]
that is, that $f(t)$ has a finite limit as $t \to \infty$. The proof of the lemma is complete.\\
\hspace*{5.5in} $\Box$