If $f$ is convex on $[a,b]$ or $[b,a]$ then
$$ f\Big(\frac{a+b}2\Big) \le \frac1{b-a}\int_a^b f(x)\,dx \le \frac{f(a)+f(b)}2 \tag{$\ast$} $$
(The first inequality is by Jensen, since $\frac{a+b}2=\frac1{b-a}\int_a^b x\,dx$; the second comes from the change of variables $x=(1-t)a+tb=a+t(b-a)$ and applying the convexity of $f$ pointwise.)
Taking $f(x)=\frac1x$ and taking reciprocals throughout yields
$$ \frac1{\frac12(\frac1a+\frac1b)} \le \frac{b-a}{\log b-\log a} \le \frac{a+b}2 $$
which is the inequality of the harmonic, logarithmic, and arithmetic means. By squeezing,
$$ \lim_{a\to b} \frac{b-a}{\log b-\log a} = b $$
Notes:
- In fact the logarithmic mean is also bounded by the geometric mean, i.e.,
$$ \sqrt{ab} \le \frac{b-a}{\log b - \log a} $$
This is stronger than the bound by the harmonic mean proved above, but the only proof I know is to take $f(x)=e^x$ in ($\ast$), and to evaluate the resulting integral we need to use $\frac{d}{dx} e^x = e^x$.
- An advantage (?) of this proof is that it doesn't need the fundamental theorem of calculus. (In fact it verifies FTC for $\int\frac1x\,dx$.) That's assuming you define $\log$ as the integral of $\frac1x$, that you prove a change of variables theorem for linear changes of variables directly by Riemann sums, and that you evaluate $\int_a^b x\,dx$ directly by Riemann sums.