Prove a relation between functions from an inequality relating limits

Question

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 $h>0$.

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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 $f(h) > g(h)$.

Answer 1

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 $|h| < \min\{\delta_1, \delta_2\}$.

Answer 2

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).