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?

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 $$\frac{f(y)-f(x)}{y-x}-f'(c)
=\frac{y-c}{y-x}S(y)+\frac{c-x}{y-x}S(x).$$
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