Prove that if $f'(x)\gt 0$ , then the function is increasing, and if $f'(x)\lt 0 $ then the function is decreasing. 
I don't really need help with both parts because the proofs are going to be mostly quite similar so I'll just ask about part (a).
So these are the things I know 
$$
f'(r)=\lim_{h->0} \frac{f(r+h)-f(r)}{h}\gt0.
$$ I also see that the problem is trying to prove that if the derivative of a function at a specific point is greater than zero, than the original function is increasing sufficiently close to that point, in this case we want to be within $\delta$ of $r$. I also notice that the interval $(r, r+\delta)$ corresponds to a right sided limit, and the interval $(r-\delta, r)$ corresponds to a left sided limit. So somehow from all these facts I basically need to deduce the inequality $f(r) \lt f(x)$ for the first interval and $f(r) \gt f(x)$ for the second interval, but I'm not totally sure how. 
This is a rough proof of the problem, but I don't know that I did it correctly
Since f ’(r) exists and is > 0, the lim as h goes to zero of [f(r+h) – f(r)]/h > 0. This means the slope of the line through f(r+h) and f(r) is positive for values of h sufficiently close to zero as long as the range is bound by f ‘(r) – ε and f ‘(r) + ε.
If we let |h| < δ and δ > 0, then r and r+h are contained within the interval (r – δ, r + δ). So since r + h, and r are both in these bounds and we have shown that the slope between f(r+h) and f(r) is positive for a value of h sufficiently close to r, then f(r) < f(x) when we pick a value of x to the right of r and yet to the left of  r+δ, and also f(r) > f(x) when we pick a value of x to the left of r and yet to the right of r - δ
 A: To do this think about what the derivative really means - that is, what is the limit, not our intuition, trying to say? In particular, the expression
$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$
says that the slope of the line from $f(x)$ to $f(x+h)$ can be constrained to within the range $(f'(x)-\varepsilon,f'(x)+\varepsilon)$ for any $\varepsilon$ by constraining $h$ to be close enough to $0$. Geometrically, this says that if we drew a line with slope a little more than $f'(x)$ and a line with slope a little less than $f'(x)$ through the point $(x,f(x))$, the function $f$ would, in some neighborhood of $x$, stay between the two lines.
The particular thing to note is that, suppose that $f'(x)>0$. Then, there must be some $\delta$ such that, if $|h|<\delta$, then $\frac{f(x+h)-f(x)}{h}>0$. This is from the definition of the limit - it represents bounding $f$ by a horizontal line (which has slope less than $f'(x)$) and some other line which, for now, we don't really care about. Of course, this implies that if $h$ is positive and within $(0,\delta)$, we get
$$f(x+h)>f(x)$$and if $h$ is within $(-\delta,0)$, we get (since when we multiply the $h$ out of the denominator, we reverse the order as $h<0$):
$$f(x+h)<f(x).$$
A: The fact that derivative exists means that for every $h >0$ there exists (we can choose it) $\varepsilon= x-x_0>0$  such that quotient
$$q=\frac{f(x_0+\varepsilon)-f(x_0)}{\varepsilon}$$
will satisfy
$f'(x_0)-h<q<f'(x_0)+h$ and $\forall x : (x_0, x)$ (Cauchy's definition of function's limit).
Therefore
$$f(x_0+\varepsilon)-f(x_0)>\varepsilon(f'(x_0)-h) ,\forall x : (x_0, x)$$
If we choose $h\leq f'(x_0)$ we are done.
