Does $f(x)\leq g(x)$ imply $\lim_{x\to 0_+} f(x)/x \leq \lim_{x\to 0_+} g(x)/x$ Let $f,g:[0,\infty)\mapsto[0,\infty)$ be convex functions such that $f(x)\leq g(x)$ for all $x\in [0,\infty).$ Suppose that both
$ \lim_{x\to 0_+} \frac{f(x)}{x}$
and
$ \lim_{x\to 0_+} \frac{g(x)}{x}$ are well defined and finite. Is it true that
$$ \lim_{x\to 0_+} \frac{f(x)}{x} \leq \lim_{x\to 0_+} \frac{g(x)}{x}?$$
It follows immediately that $ \lim_{x\to 0_+} {f(x)} \leq  \lim_{x\to 0_+} {g(x)}$ but the factor of $1/x$ in the above throws me off. Any proofs/guidance/counter examples would be appreciated.
 A: I am a student too so i hope this is correct.
Let note:$\lim_{x\to 0_+}f(x)/x=l_f\geq 0$ and $\lim_{x\to 0_+}g(x)/x=l_g \geq 0$ Because by assumption $f(x)\geq$ and $g(x)\geq0$ for $x \geq 0$
By def we can writte: $\forall \epsilon>0 , \exists\delta_f>0 \; s.t. \; \forall x\in(0; \delta_f) \Rightarrow |\frac{f(x)}{x}-l_f|<\epsilon \;or\; -\epsilon+l_f<\frac{f(x)}{x}<\epsilon+l_f$
Same for $g(x)/x$
Now let suppose that $l_f>l_g$ so by def $\exists \epsilon_0=\frac{l_f-lg}{4}>0$ s.t:
$\exists\delta_{f0}>0 \; \forall x\in(0; \delta_{f0}) \Rightarrow |\frac{f(x)}{x}-l_{f0}|<\epsilon_0$
$\exists\delta_{g0}>0 \; \forall x\in(0; \delta_{g0}) \Rightarrow |\frac{g(x)}{x}-l_{g0}|<\epsilon_0$
So we get for: $\delta_{min}=min(\delta_{f0},\delta_{g0}) \Rightarrow \forall x\in(0; \delta_{min})$  we have $|\frac{f(x)}{x}-l_{f}|<\epsilon_0$ and $|\frac{g(x)}{x}-l_{g}|<\epsilon_0$
So: $\frac{g(x)}{x}<\epsilon_0+lg=\frac{l_f+3l_g}{4}$ and $\frac{f(x)}{x}>l_f-\epsilon_0=\frac{3l_f+l_g}{4}$.
CONTRADICTION! because in this case we get: $\forall x\in(0; \delta_{min})$ the following inequality: $ \frac{g(x)}{x}<\frac{f(x)}{x} \Rightarrow g(x)<f(x)$ (as x>0).
Because: $\frac{l_f+3l_g}{4}<\frac{3l_f+l_g}{4}\Leftrightarrow l_f+3l_g<3l_f+l_g \Leftrightarrow l_f-l_g>0$. By absurd assumption we supposed $l_f>l_g$
Q.E.D
