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!

Since $F(x)/x$ tends to a limit $F'(0)>0$ as $x\to0$, there is $h>0$ so that $F(x)/x>0$ for all $x\in(0,h)$.
Thus $F(x)>0$ for all $x\in(0,h)$, and your result follows.
The condition $F(x)/x\to F'(0)$ only means that $F(x)/x$ is close to $F'(0)$ (or positive) near $0$; the limit "does not see" the values for large $x$.
What large means depends on the situation but by the definition of a limit there always is a number $h$ as above.
