= Solution
Let $D=f'_+(a)$ and write $g(t)=(f(t)-f(a))/(t-a)$ for $t>a$. By the assumed right <derivative>, choose $\delta>0$ with $|g(t)-D|<\epsilon/2$ whenever $a<t<a+\delta$. Choose $b_0\in(a,\min(b,a+\delta))$.
Apply the <mean value theorem> to $f$ on $[a,b_0]$. It gives $x\in(a,b_0)$ with $f'(x)=g(b_0)$. Both $g(x)$ and $g(b_0)$ are within $\epsilon/2$ of $D$, so
$$
\boxed{\left|\frac{f(x)-f(a)}{x-a}-f'(x)\right|
\leq|g(x)-D|+|g(b_0)-D|<\epsilon.}
$$
This proves the <endpoint secant-tangent approximation> without any <continuity> assumption on $f'$. The <continuity> required by the theorem is only that of $f$ itself.
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