Looking for a proof or a counterexample: for twice continuously differentiable curves $\vec{x}(t) \in \mathbb{R}^n$, does$$\left( \lim_{t \to \infty} \vec{x}(t) = \vec{L} \quad \text{and} \quad \lim_{t \to \infty} \vec{x}''(t) = \vec{0} \right)$$imply$$\lim_{t \to \infty} \vec{x}'(t) = \vec{0}?$$
My attempt at the problem: Neither of these conditions by themselves are enough to guarantee the result, since if we only had the first we could take $x(t) = \frac{\sin(t^4)}{t^2 + 1}$, and if we only had the second one, since $x''(t) = \frac{1}{t}$ with integral $ln(t)$ is continuous on $[1, \infty)$ and the integral over $[1, \infty)$ diverges, we should be able to smoothly interpolate this to the left, then integrate that twice to get a valid $x(t)$ counterexample in 1-D. However, I haven't been able to find a counterexample assuming both hypotheses, and I'm not sure whether the statement should be true or untrue, even just in 1-D.