For $\alpha \in (0,1)$ let the sequence $\{x_n\}$ be such that $x_0 = 0, x_1 = 1 $ and $x_{n+1} = \alpha x_n + (1-\alpha)x_{n-1},\quad n\geq1$. Find $\displaystyle \lim_{n\rightarrow \infty}x_n$.
My try:
Since $\alpha \in (0,1)\implies (\alpha-1)\in (-1,0)$
$\begin{align}|x_{n+1}-x_n|=&|\alpha -1||x_n-x_{n-1}|\\ &\vdots\\ &|\alpha -1|^n|x_1-x_0|\end{align} \implies \{x_n\} \rightarrow 0 $
Is there any alternative proof ?
Let's look for $x_n$ in the form $\lambda^n$ $$ \\\lambda^2=\alpha\lambda+(1-\alpha) \\\lambda^2-1-\alpha(\lambda-1)=0 \\(\lambda-1)(\lambda+1-\alpha)=0 $$ $=>\lambda=1,\;\alpha-1=>x_n=A+B(\alpha-1)^n\\ 0=x_0=A+B\\ 1=x_1=A+B(\alpha-1) \\B(\alpha-2)=1=>B=\frac{1}{\alpha-2}=>x_n=\frac{1}{2-\alpha}\Big(1-(\alpha-1)^n\Big)=>\lim_{n\to+\infty}{x_n}=\frac{1}{2-\alpha} $