If $x'(t)$ is bounded for $x\geq0$ show that $\lim_{t\to\infty}{x(t)}=0$ Assume that $x(t)$ is nonnegative for $t\geq0$ and $\int_0^\infty{x(t)\;dt}&lt;\infty$. If $x'(t)$ is bounded for $x\geq0$, then show that $\lim_{t\to\infty}{x(t)}=0$.




I started with contradiction taking the limit is finite and strictly positive. Then it is easy to derive the contradiction. But the limit may be oscillatory too or diverging to $+\infty$.

Any help would be apreciated. Thanks in advance.
If $x(t)\not\to 0$, then there exists a $\delta&gt;0$ such that there exists a sequence $t_n\to\infty$ with $x(t_n)\ge \delta$. Taking a subsequence we can assume $t_{n+1}-t_n\ge \frac\delta M$. We have $|x'(t)|\le M$ so $x(t)\ge \frac\delta2$ for $|t-t_n|&lt;\frac\delta{2M}$ and so $$\int_{t_n}^{t_{n+1}} x(t)dt \ge \frac{\delta^2}{4 M}.$$ Hence the integral must be infinite.
Incomplete proof (missing the case when $x(t)$ has no limit)

By contradiction, let $L \gt 0$ such that $\lim_{t \to \infty} x(t) = L$.

Then, by definition $\forall$ $\varepsilon \gt 0$ exists $k \in \mathbb{R^+}$ such that
$$\mid x(t) - L \mid \lt \varepsilon \;\;\;\;\;\;\;\; \forall \;\;t \geq k$$
Let $\varepsilon \in (0, L)$, then
$$0 \lt L - \varepsilon \lt x(t) \;\;\;\;\;\;\;\; \forall \;\; t \geq k$$
$$\Rightarrow 0 &lt; \int_k^\infty (L - \varepsilon)dt &lt; \int_k^\infty x(t)dt &lt; \infty$$
But this is a contradiction since $L - \varepsilon$ is a positive constant so $\int_k^\infty (L - \varepsilon)dt$ is not bounded.