Can someone please explain how to supply a rigorous $\epsilon,\delta$ argument for this theorem as Spivak says ?
My argument is:
$f'(a)=\lim_{h\to 0} f'(\alpha_h)$ equivalent to $|f'(\alpha_h)-f'(a)|<\epsilon$ if $0<|h|<\delta$.
Since $\alpha_h$ takes on every value in the interval $(a,a+h)$ and approaches $a$ as $h\to 0$, thus $\alpha_h=x=a+h$.
So $|f'(x) -f'(a)|<\epsilon$ if $0<|x-a|<\delta$ which is equivalent to $\lim_{x\to a}f'(x)=f'(a)$.
Can someone please explain how the correct argument would be ?