Is there a real-analytic monotone function $f:(0,\infty) \to \mathbb{R}$ which vanishes at infinity, but whose derivative admits no limit?

What we need is a real-analytic non-negative and integrable $g$ that has no limit at $+\infty$. Then $$f(x) = \int_x^{+\infty} g(t)\,dt$$ fits the bill.

Consider $$g(x) = \biggl(\frac{2 + \cos x}{3}\biggr)^{6 x^5}\,.$$

It is evident that $g$ is strictly positive, real-analytic on $(0,+\infty)$, and has no limit as $x \to +\infty$. It remains to see that $g$ is integrable. For a positive integer $n$, consider the interval of length $\pi$ with midpoint $n\pi$. In this interval, for $\lvert x - n\pi\rvert \geqslant \frac{1}{n^2}$ we have $$\lvert \cos x\rvert \leqslant \cos \bigl(n^{-2}\bigr) \leqslant 1 - \frac{1}{2n^4} + \frac{1}{24n^8} \leqslant 1 - \frac{1}{3n^4}$$ by Taylor expansion, and hence (using $\bigl(n - \frac{1}{2}\bigr)\pi > \frac{3}{2}n$) $$g(x) \leqslant \biggl(1 - \frac{1}{9n^4}\biggr)^{9n^5} \leqslant \exp \bigl(-n\bigr)\,.$$ Hence the integral of $g$ over that interval is bounded by $$\frac{2}{n^2} + \pi\cdot e^{-n}\,,$$ which is a summable sequence.