Prove a relation between functions from an inequality relating limits Given two functions $f$ and $g$ satisfying the inequality $$\lim_{h \to 0} \frac{f(h)}{h} > \lim_{h \to 0} \frac{g(h)}{h} \tag{1},$$
prove that there exists $h>0$ such that $f(h) > g(h)$. I was thinking that intuitively, if $f(0) = g(0)$ and $f'(0) > g'(0)$, $f(h) > g(h)$ for some $h>0$.

I have tried the following using the epsilon-delta definition of limits.
Define the following quanities
\begin{align}
A = \lim_{h \to 0} \frac{f(h)}{h} \tag{2}\\
B = \lim_{h \to 0} \frac{g(h)}{h} \tag{3}
\end{align}
where $A>B$. For any $\epsilon > 0$, there exists $\delta_1$ and $\delta_2$ such that
\begin{align}
|h| < \delta_1 \implies \left| \frac{f(h)}{h} - A\right| < \epsilon \tag{4}\\
|h| < \delta_2 \implies \left| \frac{g(h)}{h} - B\right| < \epsilon \tag{5}
\end{align}
From equation $(4)$ and $(5)$, I derived the following results for $h>0$:
\begin{align}
A &< \frac{f(h)}{h} + \epsilon \tag{6} \\
B &> \frac{g(h)}{h} - \epsilon \tag{7} \\
A > B \implies f(h) &> g(h) - 2 \epsilon h \tag{8}
\end{align}
However, I am unsure how to proceed from equation $(8)$ to get $f(h) > g(h)$.
 A: Intuitively, you have two intervals,
$$\left(A - \varepsilon, A + \varepsilon\right) \text{ and } \left(B - \varepsilon,B + \varepsilon\right),$$
and if you take $|h|$ sufficiently small (in your case, if you take $|h|<\delta_1$ and $|h|<\delta_2$),
$$\frac{f(h)}{h} \in \left(A - \varepsilon, A + \varepsilon\right) \text{ and } \frac{g(h)}{h} \in\left(B - \varepsilon,B + \varepsilon\right).$$
We are done if the two intervals were disjoint; if so,
$$\frac{f(h)}{h} > A-\varepsilon \geq B+\varepsilon > \frac{g(h)}{h}.$$
But we are free to choose $\varepsilon$. In this case
$$A-\varepsilon \geq B + \varepsilon \iff \varepsilon \leq \frac{A-B}{2}$$
so just take such one such $\varepsilon>0$ (then we have the necessary $\delta_1$, $\delta_2>0$, so take any $|h| < \min\{\delta_1, \delta_2\}$.
A: This is a simpler but different approach. Let $f(0)=g(0)=0$ then $f, g$ are continuous and differentiable at $0$ and by the conditions of the question $\phi'(0)>0,\phi(0)=0$ where $\phi(x) =f(x) - g(x) $. Clearly this means that $\phi$ is strictly increasing at $0$ and hence there are numbers $h, k$ both greater than $0$ such that $\phi(x) > \phi(0)=0$ for all $x\in(0,h)$ (this is what the question asks us to prove) and further $\phi(x) <\phi(0)=0$ for all $x\in(-k, 0)$ (this is bonus).
