Can we derive an exponential bound from a special limit of a function?

Question

Assume $\lambda>0$ and a continous function $x(t), x:[0,\infty) \rightarrow \mathbb{R}^n$. Does it hold that $$ \lim_{t \to \infty}\frac{1}{t} \log(|x(t)|)<0 \Leftrightarrow \exists C>0: \forall t\geq 0 \quad |x(t)| \leq C\cdot\exp^{-\lambda t} $$ ? From right to left is trivial, for the opposite direction I'm unsure wheter it holds.
This problem comes from different definitions for stability, when the statement holds, then they are equivalent.

Answer 1

If $$ \lim_{t\to\infty}\frac{1}{t} \log(|x(t)|)<0$$ then there exist positive numbers $T$ and $\lambda$ such that $\frac{1}{t} \log(|x(t)|)<-\lambda $ for $t >T$. So $|x(t)|<e^{-\lambda t}$ for $t>T$. Now use the fact that $e^{\lambda t} x(t)$ is continuous, hence bounded on $[0,T]$.