Spivak Ch 11 Theorem 7

Question

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 ?

Answer 1

Let $\displaystyle L = \lim_{x \to a} f'(x)$ and let $\epsilon > 0$ be given.

We may select $\delta > 0$ with the property that $0 < |x-a| < \delta$ implies $|f'(x) - L| < \epsilon$.

Now, if $0 < |x-a| < \delta$ we may invoke the mean value theorem to find a point $c$ in between $a$ and $x$ satisfying $\dfrac{f(x) - f(a)}{x-a} = f'(c)$. Note that $c$ also satisfies $0 < |c-a| < \delta$, so that $|f'(c) - L| < \epsilon$. Consequently $\left| \dfrac{f(x) - f(a)}{x-a} - L \right| < \epsilon$. In other words, $$0 < |x-a| < \delta \implies \left| \frac{f(x) - f(a)}{x-a} - L \right| < \epsilon.$$

This is precisely what $f'(a) = L$ means.