Let $f_\lambda :[0,1] \to [0,1]$ be the logistic map with parameter $1 < \lambda < 3$, $$f_\lambda(x) = \lambda x (1-x).$$ The two fixed points of $f_\lambda$ are $0$ and $a = 1 - \frac{1}{\lambda}$.
Prove that for every $x \in (0,1)$, the orbit $\{f^{\circ k}_\lambda(x) \}_{k=1}^\infty$ converges to $a = 1 - \frac{1}{\lambda}$.
This fact is claimed in Chapter 13 of Falconer's Fractal Geometry textbook (page 174). I am trying to show it analytically.
After choosing an initial $x_0 \in (0,1)$, we generate the iterates:
$$x_1 = \lambda x_0 (1 - x_0), x_2 = \lambda x_1 (1 - x_1), \dots .$$
If we write $$x_k = a + \delta_k ,\quad \frac{1}{\lambda} - 1 \le \delta_k \le \frac{1}{\lambda}, $$
then using Taylor's formula we get
$$\delta_{k+1} = (2 - (1+\delta_k)\lambda)\delta_k$$
But now it's not clear to me how to use this to obtain convergence. We would like to have $|(2 - (1+\delta_k)\lambda)|$ very small to use a contraction argument. But, for instance, if $x_0 = 0.9$ and $\lambda = 2.7$, then $(2 - (1+\delta_0)\lambda) = -1.43$. Also the fact that the $\delta_k$'s are signed seems to cause issues. So it seems some particular cases need to be analyzed individually before one proves the entire result.
Any hints or solutions are greatly appreciated!