How to 'get rid of' limit so I can finish proof? Suppose $\sup_{x \in \mathbb{R}} f'(x) \le M$.
I am trying to show that this is true if and only if $$\frac{f(x) - f(y)}{x - y} \le M$$
for all $x, y \in \mathbb{R}$.
Proof
$\text{sup}_{x \in \mathbb{R}} f'(x) \le M$
$f'(x) \le M$ for all $x \in \mathbb{R}$
$\lim_{y \to x} \frac{f(y) - f(x)}{y - x} \le M$
$\lim_{y \to x} \frac{f(x) - f(y)}{x - y} \le M$
I can see geometrically why this property holds, but how do I get rid of the limit here? Or am I approaching it wrong in general?
 A: The 'if' part follows from the definition of the derivative as a limit. If some expression is always less than or equal to $M,$ and the limit exists, then the limit also satisfies that inequality. That, in its turn, follows from the epsilon-delta definition of a limit.
The 'only if' part is the really interesting part. As commenters have pointed out it is (a direct consequence of) the Mean Value Theorem.
A: well, first of all, we have to presume f is continuous and differentiable on R.  This statement isn't true otherwise.
1)  Suppose $\frac{f(x) - f(y)}{x - y} > M$ for some $x, y \in \mathbb R$.
By the mean value theorem, there exist a $c; x <c < y$ where $f'(c) = \frac{f(x) - f(y)}{x - y}$.
So $f'(c) > M$.
So $\sup f'(x) \le M \implies f'(c) \le M$ for all $c \in \mathbb R \implies$ $\frac{f(x) - f(y)}{x - y} \le M$ for all $x, y \in \mathbb R$.
2) Suppose $\frac{f(x) - f(y)}{x - y} \le M$ for some $x, y \in \mathbb R$.
Then  $\lim_{y \rightarrow x}\frac{f(x) - f(y)}{x - y} = f'(x) \le M$ for all $x \in \mathbb R$.  So {$f'(x)|x \in \mathbb R$} is bounded above by M so $\sup_{x \in \mathbb{R}} f'(x) \le M$.
