If $f$ increasing, analytic on $\mathbb{R}$ and $\lim_{x\to +\infty}f(x)=1$, does it follows that $\lim_{x\to +\infty}f'(x)=0$?

$$ f(x) = \int_0^x \exp(-t^4 \sin^2(t))\, dt $$

is a counterexample, after normalizing by a suitable constant so that $ \lim_{x \to \infty} f(x) = 1 $ instead of $ 1.17195 \ldots $

The idea is that the factor $ t^4 $ suppresses the contribution of the integrand at all points too far from the zeroes of $ \sin(t) $ as $ t $ gets large, and $ \sin^2(t) $ is only within $ O(1/N^4) $ of $ 0 $ for $ t $ that's $ O(1/N^2) $ from an integer multiple of $ \pi $. Since $ 1/N^2 $ is a fast enough decaying sequence, the integral converges. It's obviously an analytic function.

Edit: Here is a more formal argument. Consider the integral

$$ \int_{N \pi}^{(N+1) \pi} \exp(-t^4 \sin^2(t)) \, dt \leq \int_0^{\pi} \exp(-\pi^4 N^4 \sin^2(t))$$

Split the interval $ [0, \pi] $ into two pieces, $ [1/N^{5/4}, \pi - 1/N^{5/4}] $ and its complement. Over this interval,

$$ \sin^2(t) \geq \sin^2(1/N^{5/4}) = 1/N^{5/2} + O(1/N^5) $$

so that $ \exp(-\pi^4 N^4 \sin^2(t)) \leq \exp(-\pi^4 N^{3/2} + O(N^{-1})) $, which is obviously summable by comparison to the geometric series, say. So the contribution to the integral from this part is $ L^1 $. The other part of the integral is over a set of measure $ 2/N^{5/4} $, and since the integrand is bounded from above by $ 1 $, this part only contributes a term of order $ O(1/N^{5/4}) $, which is also $ L^1 $. We conclude therefore that $ f $ is well-defined by monotonicity and has a finite limit at infinity.