Question about derivatives and inequalities Problem: Assume that $f: \mathbb R\rightarrow\mathbb R$ and $g: \mathbb R\rightarrow\mathbb R$ are differentiable and $f(0) = g(0)$ and $f'(0) < g'(0)$. Prove that there exists $h > 0$, so that $f(x) < g(x)$ when $0<x<h$.
I tried to solve this by letting $F(x) = g(x) - f(x)$, where $F(0) = g(0) - f(0) = 0$ and $F'(0) = g'(0) - f'(0) > 0$. Also due to the definition of derivative
$F'(0) = \lim_{x\to0}$ ${F(x) - F(0)\over x-0}$ = $\lim_{x\to0}$ $F(x)\over x$
and as we assumed $F'(0) > 0$, so must be $F(x) > 0$ if $x>0$. This leads to $g(x) - f(x) > 0$ and thus $g(x) > f(x)$ when $x>0$, which doesn't seem right because I can't show that the required $h>0$ exists. Any help would be appreciated!

As Joonas says, the limit gives us information only when $x$ is close to $0$. In particular, we know that
$$\forall \epsilon>0,\ \exists\delta>0 \quad\text{ such that }\quad 
0<|x|<\delta\ \Longrightarrow\ \left|\frac{F(x)}{x}-F'(0)\right|<\epsilon$$
Since $F'(0)>0$, when $\epsilon < F'(0)$, we can find $\delta$ such that
$$0<|x|<\delta\ \Longrightarrow\ \frac{F(x)}{x}>0$$
In fact, we can consider the cases $x>0$ and $x<0$:
\begin{align*}
0<x<\delta\ \Longrightarrow\ F(x)>0\\
-\delta<x<0\ \Longrightarrow\ F(x)<0
\end{align*}
And we can conclude

\begin{align*}
f(x)<g(x)\quad &\text{if}\quad\quad\  0<x<\delta\\
f(x)>g(x)\quad &\text{if}\quad -\delta<x<0
\end{align*}

