advanced calculus, differentiable and limit problem Let $f:[a,b]\rightarrow R$ be differentiable at $c\in [a,b]$. Show that for every $\epsilon >0$, there is a $\delta(\epsilon) >0$ s.t if $0<|x-y|<\delta(\epsilon)$ and $a\leq x \leq c\leq y \leq b$, then\
$$ |{\frac{f(x)-f(y)}{x-y}-f'(c)}|<\epsilon$$
I can only think of using triangle inequality, but it seems does not work. Can anyone help me?
 A: Hint: If $x=c<y$ or $x<c=y$ this is immediately immediate from the definition. So assume $x<c<y$. Then $$\frac{f(y)-f(x)}{y-x}-f'(c)
=\frac{y-c}{y-x}\left(\frac{f(y)-f(c)}{y-c}-f'(c)\right)+\frac{c-x}{y-x}\left(\frac{f(c)-f(x)}{c-x}-f'(c)\right).$$
Now you can use the triangle inequality there; when you use the triangle inequality it's going to be very important that the two numbers $(y-c)/(y-x)$ and $(c-x)/(y-x)$ are both positive and add up to $1$.
Added: Of course user254665's answer is cleaner, applying directly to the case $x\le c\le y$. It seems possible that my version will seem simpler to some readers (and that his will seem simpler to others). A compromise, a version of my version that doesn't require handling the cases $x=c$ and $y=c$ separately:
Define $$S(x)=\begin{cases}\frac{f(x)-f(c)}{x-c}-f'(c),&(x\ne c),
\\0,&(x=c).\end{cases}$$ Then for $x\le c\le y$ we have $$\frac{f(y)-f(x)}{y-x}-f'(c)
=\frac{y-c}{y-x}S(y)+\frac{c-x}{y-x}S(x).$$
[...]
A: $$\text {Let }  d>0 \text { where }  0<|z-c|<d \implies |\frac {f(z)-f(c)}{z-c}-f'(c)|<e/3.$$ For $c-d<x\leq c\leq y<c+d)$ with $x\ne y$ we have $$f(x)=f(c)+(x-c)(f'(c)+e_x),$$ $$ f(y)=f(c)+(y-c)(f'(c)+e_y),$$ $$\text {where }\; |e_x|<e/3  \;\text { and }\; |e_y|<e/3.$$ $$\text {Therefore }\; f(x)-f(y)=(x-y)f'(c)+(x-c)e_x-(y-c)e_y=$$ $$=(x-y)(f'(c)+(x-c)(e_x-e_y)+e_y(x-y).$$ $$\text {Thus }\; |\frac {f(x)-f(y)}{x-y}-f'(c)|=|(e_x-e_y)\frac {(x-c)}{(x-y)}+e_y|\leq $$ $$\leq (|e_x|+|e_y|)\frac {|x-c|}{|x-y|}+|e_y|\leq (|e_x|+|e_y|)+|e_y|<e$$ $$\text {because }\; |x-c|/|x-y|\leq 1.$$
