Limes superior and inferior inequalities with functions and its derivatives

Question

Lemma 2. from N.H. Duu, On The Existence of Bounded Solutions for Lotka- Volterra Equations, Acta Mathematica Vietnamica 25(2) (2000), 145-159.

Let $G(t)$ and $F(t)$ be two diferentiable functions defined on $(0,\infty)$ such that $\lim_{t\to\infty} G(t) = \lim_{t\to\infty}F(t) = +\infty$ then $$\limsup_{t\to\infty}\frac{G(t)}{F(t)}\leq \limsup_{t\to\infty}\frac{G'(t)}{F'(t)}; \quad\liminf_{t\to\infty}\frac{G(t)}{F(t)}\geq \liminf_{t\to\infty}\frac{G'(t)}{F'(t)}.$$

Proof. By the Cauchy theorem for diferential functions, for any $t_{1}, t_{2} > 0$ there is an $\theta \in (t_{1}, t_{2})$ such that $$\frac{G'(\theta)}{F'(\theta)}=\frac{G(t_{1})-G(t_{2})}{F(t_{1})-F(t_{2})}=\dfrac{G(t_{2})}{F(t_{2})}\times\dfrac{1-\dfrac{G(t_{1})}{G(t_{2})}}{1-\dfrac{F(t_{1})}{F(t_{2})}}.$$ Letting $t_{1}$ and $t_{2} \to \infty$ such that $\lim\dfrac{G(t_{1})}{G(t_{2})} = lim \dfrac{F(t_{1})}{F(t_{2})}= 0$ we get the result.

Can anybody help me understand the last part of this proof? Exactly, how this transformation $$\frac{G'(\theta)}{F'(\theta)}=\frac{G(t_{1})-G(t_{2})}{F(t_{1})-F(t_{2})}=\dfrac{G(t_{2})}{F(t_{2})}\times\dfrac{1-\dfrac{G(t_{1})}{G(t_{2})}}{1-\dfrac{F(t_{1})}{F(t_{2})}} \text{where} \lim_{t_1,t_2\to\infty}\dfrac{G(t_{1})}{G(t_{2})} = \lim_{t_1,t_2\to\infty} \dfrac{F(t_{1})}{F(t_{2})}= 0$$ leads to inequalities from lemma?

Answer 1

The cited proof lacks many details. So I expand the limit superior version. The inferior version is analogous.

1. Since $F(t) \to \infty$ for $t\to\infty$ we may assume that $F > 0$ without lost of generality.

2. Fix some $t_1\in (0,\infty)$. We may assume that for every $t > t_1$ it follows $F(t) > F(t_1)$.

3. By Cauchy mean value theorem, for each $t_2 > t_1$ there exists some $\theta \in (t_1, t_2)$ with $$ \frac{G(t_2) - G(t_1)}{F(t_2)} \le \frac{G(t_2) - G(t_1)}{F(t_2) - F(t_1)} = \frac{G'(\theta)}{F'(\theta)} \le \sup_{t > t_1} \frac{G'(t)}{F'(t)} =: M(t_1). $$

4. In particular for $t_2 > t_1$, we have $$ \frac{G(t_2)}{F(t_2)} = \frac{G(t_2) - G(t_1)}{F(t_2)} + \frac{G(t_1)}{F(t_2)} \le M(t_1) + \frac{G(t_1)}{F(t_2)}. $$

5. Compute the limit superior with respect to $t _2 $: $$ \limsup_{t_2\to\infty} \frac{G(t _2)}{F(t _2)} \le M(t_1) + 0. $$ Notice that the left-hand-side becomes independent of $t_1$.

6. Now apply the limit with respect to $t_1$: $$ \limsup_{t _2\to\infty} \frac{G(t _2)}{F(t _2)} \le \lim_{t_1\to \infty} M(t_1) = \limsup_{t\to\infty} \frac{G'(t)}{F'(t)}. $$ And we are done.