Existence of continuous $r(t)$ with $\lim_{t \to \infty} \frac{f(r(t))}{g(t)} = 1$ Let $ \ f,g: \mathbb{R} \to \mathbb{R} \ $ be continuously differentiable functions such that $$\lim_{t \to \infty} f(t) = \infty = \lim_{t \to \infty} g(t) \ \ . $$ My question is:

Is there a continuous function $ \ r: \mathbb{R} \to \mathbb{R} \ $ such that $$\lim_{t \to \infty} \frac{f \big( r(t) \big)}{g(t)} = 1 \ \ \ \ ? $$

I would like hints for a proof or a counterexample. You may feel free to modify the assumptions about $f$ and $g$ as you please.
Thanks in advance.

Consider the functions $f(t)=t(2+\cos t)$ and $g(t)=t$. They are infinitely differentiable at the whole domain. Now suppose that the above function $r$ exists.
Let $T>0$ be large enough for $\frac 2 3 < \frac {f(r(t))}{g(t)} < \frac 4 3$ to hold for all $t>T$. This implies $$T<t_1<t_2 \implies \frac {f(r(t_1))}{f(r(t_2))} < \frac {\frac 4 3 \, g(t_1)}{\frac 2 3 \, g(t_2)}=2\,\frac {t_1}{t_2} < 2\,. \tag{1}\label{eq1}$$
Note that $|f(t)| \le 3|t|$, $f(t)$ has the sign of $t$, hence for all $t>T$ we have: $\frac 2 3 t = \frac 2 3 g(t) < f(r(t)) \le 3 r(t)$ and therefore $\lim\limits_{t \to +\infty} r(t) = +\infty$. Select $n \in \mathbb N$ such that $2 \pi n > r(T)$ and let $$t_1=\inf\,\{t>T \mid r(t)=2 \pi n\},\quad t_2=\inf\,\{t>T \mid r(t)=(2n+1) \pi\}.$$ By the intermediate value theorem, $T<t_1<t_2$. Now we see that $$\frac {f(r(t_1))}{f(r(t_2))}=\frac {f(2\pi n)}{f((2n+1) \pi)}=\frac {2 \pi n \cdot 3}{(2n+1) \pi \cdot 1}=\frac {6n}{2n+1} \ge 2\,,$$ that contradicts $\eqref{eq1}$.
