differentiable and uniform continuity of f and F Given $f: \Bbb R \to \Bbb R$. define new function: $F(x) =\frac{f(x)-f(a)}{x-a}$ for $x\neq a$. Prove that $f$ is differentiable at $a$ if and only if $F$ is uniformly continuous in some punctured neighborhood around $a$.
I was playing around with the definitions of both and I think the backward direction is easier but I'm having trouble with the absolute values which come with the definitions.
I was thinking of defining $F(x)$ to be $f'(a)$ since it is a limit definition of the derivative. Am I on the right track?
 A: The forward direction is false.
Let $f(x) = x^2 1_\mathbb{Q}(x)$, then $f$ is differentiable at $x=0$, but
$F(x) = x 1_\mathbb{Q}(x)$, while continuous at $x=0$ is not continuous 
at $x$ for any $x \neq 0$.
The reverse direction is straightforward:
Suppose $f$ is differentiable at $a$ and define $F(a) = f'(a)$. Then let $\epsilon>0$ and choose $\delta>0$ such that
$|f(x)-f(a) -f'(x) (x-a)| \le {1 \over 2} \epsilon |x-a|$ for all $x \in B(a, \delta)$.
Now suppose $x,y \in B(a,\delta)$, with $\delta \le 1$, then $|f(x)-f(y)| \le |f(x)-f(a) -f'(x) (x-a)|+ |f(y)-f(a) -f'(y) (y-a)| \le \epsilon $, and
so $F$ is 
Suppose $F$ is uniformly continuous in $B(a,r)\setminus \{x\}$. Since $F$ is
uniformly continuous, we can uniquely extend the domain of $F$ to include $a$.
Let $\epsilon>0$ and choose $\delta>0$ such that $|F(x)-F(a)| < \epsilon$ for all $|x-a| < \delta$. Then, for $0<|x-a|< \delta$, we have
$| { f(x)-f(a) \over x-a} -F(a)| < \epsilon$. Hence
$\lim_{x \to a} {f(x)-f(a) \over x-a} =F(a)$ and so $f$ is differentiable at $x$
with derivative $F(a)$.
