2

If $\lim_{t \to \infty} \dot{x}(t)=0$ and $|\dot{x}(t)|$ exponentially converges to $0$ (i.e., there exist the positive constants $c_1$ and $c_2$, such that $|\dot{x}(t)|\leq c_1e^{-c_2~t}$, $t\geq 0$), can we prove the existence of $\lim_{t\to\infty}x(t)$?

CUG
  • 55
  • 1
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jul 26 '22 at 01:18
  • 1
    It is time to unleash the power of the fundamental theorem of calculus. This will give you that $(x(n))_{n\in \mathbb{N}}$ is Cauchy. Then finish with another application of the fundamental theorem of calculus. Pat your back and have a soda :) – Severin Schraven Jul 26 '22 at 01:31
  • 1
    hell have a beer if you like – Robert Lewis Jul 26 '22 at 01:41
  • Thanks Severin Schraven for your hints. Can you please explain more? – CUG Jul 26 '22 at 01:46
  • 1
    @CUG Something like this maybe: $| x(t) - x(s) | = | \int_s^t \dot{x}(r) , dr | \leq \int_s^t |\dot{x}(r)| , dr$. – Charles Hudgins Jul 26 '22 at 02:07
  • @ Charles Hudgins Thanks a lot! I know how to prove it. We can first prove $\int_{0}^{+\infty}|\dot{x}(t)|dt<\infty$ (absolute convergence), which implies $\int_{0}^{+\infty}\dot{x}(t) dt<\infty$. Thus, $x(t)$ is convergent as $t\to\infty$. – CUG Jul 26 '22 at 02:26
  • @CUG In order to ping somebody, you should not leave white space between @ and the name :) your proof works nicely. In fact even better, then what I had in mind. – Severin Schraven Jul 26 '22 at 02:33
  • @Severin Schraven Thank you very much:) – CUG Jul 26 '22 at 02:36

0 Answers0