For a $C^1$ function, the difference $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |$ is small when $|d-c|$ is small 
Suppose $g\in C^1 [a,b]$. Prove that for all $\epsilon > 0$, there is $\delta > 0$ such that $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{< \epsilon }$ for all points $c,d \in [a,b]$ with $0 <|d-c|< \delta$

First, I don't know what $C^1 [a,b]$ means.
Some ideas:
By Mean value theorem, $g'(c) =  {{g(b)-g(a)} \over {b-a}} $ since $c\in[a,b]$. 
To show $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{< \epsilon }$ whenever $0 <|d-c|< \delta$ for $ c,d \in [a,b]$. I guess I have to use the definition of limit of continuous function. But, I don't know how to connect all of these ideas. 
 A: This problem has nothing to do with the mean value theorem. This problem is designed to test your understanding of uniform continuity and/or compactness. It seems likely to me that you have just learned (in the context of whatever book or class this is from) the compactness argument that a continuous function on a compact interval is uniformly continuous.
Once you've parsed this and your question, the answer amounts to writing down a definition of uniform continuity.
A: Start simpler: Suppose that there is a function $g:[a, b] \to \mathbb{R}$ (resp. $\mathbb{C}$), differentiable at $c \in (a, b)$.  Consider the function $h_c: [a, b] \to \mathbb{R}$ (resp. $\mathbb{C})$ given by
$$h_c(d) := \begin{cases} \displaystyle \frac{g(d) - g(c)}{d - c}, & d \neq c\\
g^{\prime}(c), & d = c.\end{cases} $$


*

*Is it clear why $h_c$ is continuous at $d = c$?

*If in addition, $g$ was continuous on $[a, b]$, do you see how $h_c$ would be continuous on $[a, b]$?

*Finally, is it clear how such an expression is "natural" given the absolute-value given to you (or by just thinking about derivatives)? 


In your problem, then, you are asked to show that with a stronger assumption on $g$, $h_c$ is somehow "still nice" when $c$ and $d$ both vary; i.e., a better notation would be $h(d; c)$ or somesuch. 
Does this help?
