If $f'(x_0)>0$, how do I explicitly show there is a neighborhood of $x_0$ in which the difference quotient is strictly positive? I am reading a proof stating if $f'(x_0) >0$, then $f$ is strictly increasing at $x_0$. (from The Way of Analysis by Strichartz)
In the proof it says:
"Since the limit of the difference quotient is strictly positive, there must be a neighborhood of $x_0$ in which the difference quotient is strictly positive."
I see this if I think about a graph where the tangent line has a positive slope at a point, but I am having trouble explicitly writing it down on paper.

If $f'(x_0)>0$, how do I explicitly show there is a neighborhood of $x_0$ in which the difference quotient is strictly positive?

 A: Hint: This might make it clearer: Assume $\lim_{x\to x_0} g(x) = L > 0.$ Do you see why there exists $\delta > 0$ such that $g>0$ on $(x_0-\delta,x_0) \cup (x_0, x_0+\delta)?$
A: Hint of zhw:
$\lim_{ x \rightarrow x_0} \dfrac{f(x) -f(x_0)}{x-x_0} = f'(x_0)=: L>0:$
Let $\epsilon \gt 0$.
There is  a $\delta$ such that 
$0<|x-x_0|<\delta$ implies 
$|\dfrac{f(x)-f(x_0)}{x-x_0} - L| < \epsilon.$
Choose $\epsilon \lt L$.
A: Assume that no such neighborhood exists. Then for every neighborhood of $x_0$, there is at least one point such that the difference quotient is non-positive. Taking this sequence to $x_0$, we find a sequence of $x_i$ s.t. $\lim\limits_{i\to\infty} \frac{f(x_i)-f(x_0)}{x_i-x_0}\le 0$. But the definition of a limit is that every sequence converges to the same point, so this contradicts our assumption that the limit of the difference quotients is positive.
A: Beware: if the derivative of a function is positive it doesn’t mean that the function is monotone: Consider $f\colon\mathbb R\to\mathbb R$ defined by $f(x):=x$ for rational $x$ and $f(x):=\tan(x)$ otherwise.  Then $f$ is differentiable in $x_0=0$ with $f’(x_0)=1$ but $f$ is not monotone.
A: The meaning of the statement in italics in question should be clear if you understand the definition of limit.
You are given that $f'(x_{0})>0$ and by definition of derivative you know that $f'(x_{0})$ is the limit of the difference quotient $(f(x) - f(x_{0}))/(x-x_{0})$ as $x\to x_{0}$. Thus we know the difference quotient tends to a positive limit as $x\to x_{0}$. By definition of limit it is ensured that values of the difference quotient are as near to its limit $f'(x_{0})$ as we want by choosing a suitable neighborhood of $x_{0}$. And then comes the obvious part.

If you take values near a positive number those values will also be positive (or to be precise if a number $k$ is positive we can find a neighborhood of $k$ which contains positive numbers only, the obviousness of this fact is not affected in any way by the use of the jargon neighborhood).

Thus it is guaranteed that the values of difference quotient are positive in some suitable neighborhood of $x_{0}$.
You should also note that the result does not work if $f'(x_{0})=0$ because of the trivial fact that any neighborhood of $0$ necessarily contains both positive and negative values and thus we don't have any guarantee about the sign of difference quotient in any neighborhood of $x_{0}$.
Many proofs in analysis make use of the trivial fact mentioned above and they may seem difficult only because of the jargons or Greek symbols $(\epsilon, \delta) $ involved. Try to learn the meaning of jargons in terms of everyday language and treat the Greek symbols at par with Roman symbols $a, b, c$. 
A: This is my answer which I wrote in the comment of zhw.'s answer:

We know $$\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0) >0.$$
By definition of limit, there exists $\delta$ such that
$$0<|x-x_0|<\delta \implies \left|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)\right|<\frac{f'(x_0)}{2}.$$
Expanding the absolute value, 
$$-\frac{f'(x_0)}{2}<\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)<\frac{f'(x_0)}{2}.$$
Therefore,
$$0<\frac{f'(x_0)}{2}<\frac{f(x)-f(x_0)}{x-x_0}<\frac{3f'(x_0)}{2}.$$
So, $$0<\frac{f(x)-f(x_0)}{x-x_0}.$$
