Can a derivative of $ x $ at ${\xi}$ be thought of as a function of $ x $? It can be shown that if limit of $f(x)$ as $x$ aproaches ${\xi}$ is greater than zero, than there exists such $h$ greater than zero that $∀x:ξ−h<x<ξ+h,x≠ξ:f(x)>0$ (you can see the proof here).
Apparently it follows from the above that if $f'({\xi})>0$ there exists such an open interval that 
\begin{align}
\ ({\xi}-h..{\xi}+h),x≠ξ:\frac{f(x)-f({\xi})}{x-{\xi}}>0
\end{align}
How? Is it because $\frac{f(x)-f({\xi})}{x-{\xi}}$ allows itself to be thought of as a function of $x$ itself and therefore is subject to the same laws at its limit, or is it more complicated?
 A: Here is an easier way to understand why if the limit of $f(x)$ as $x \to a$ is positive, then $f(x)$ is positive on some neighborhood $(a - \delta, a + \delta)$ around $a$.
Let's call the limit of $f(x)$ as $x$ approaches $a$ the letter $L$ so that it's easy to talk about/reference.
Suppose $L > 0$.  Suppose on the otherhand that for every $\delta > 0$, we can find some point $y$ in $(a - \delta, a + \delta)$ such that $f(y) \leq 0$.  So for every neighborhood around $a$, we can find a point in that neighborhood that such that $f(x)$ is not positive at that point.
But if $L$ is the limit of $f(x)$ as $x \to a$, then by definition that means for each $\epsilon > 0$, we can find a $\delta > 0$ such that if $x$ is in $(a - \delta, a + \delta)$, then $f(x)$ is in $(L - \epsilon, L + \epsilon)$.  Since this is true for every $\epsilon > 0$, we can shrink $\epsilon$ small enough so that $(L - \epsilon, L + \epsilon)$ contains only positive points.  Then even in this interval, by the definition of limit, we should be able to find some $\delta > 0$ such that for all elements in $(a - \delta, a + \delta)$, their images under $f$ are all in $(L - \epsilon, L + \epsilon)$.  
But this contradicts what we said earlier that for each $\delta > 0$, we can find a point $y$ in $(a - \delta, a + \delta)$ such that $f(y) \leq 0$, because we can't have $f(y) \leq 0$ and $f(y) \in (L - \epsilon, L + \epsilon)$ at the same time.
So, either the limit is not $L$, or if it is $L$, then for some $\delta > 0$, all points in $(a - \delta, a + \delta)$ will have image in $(L - \epsilon, L + \epsilon)$ (and thus will have positive image).
