The question is :
Suppose ${\{x_n\}_{n=1}^\infty}$ is a sequence, ${x \in \mathbb{R}}$, and ${x_n \ne x}$ for all ${n \in \mathbb{N}}$. Suppose the limit
L = ${ \lim_{n\to \infty}}$ ${\dfrac{|x_{n+1} - x|}{|x_n - x|}}$
exists and L $\lt$ 1. Show that ${\{x_n\}_{n=1}^\infty}$ converges to x.
The problem I faced :
My intuition says to use ratio test as here ${\dfrac{|x_{n+1}- x|}{|x_n - x|}}$ form is given.
From the given information i.e, L = ${ \lim_{n\to \infty}}$ ${\dfrac{|x_{n+1} - x|}{|x_n - x|}}$,
we get,
${ \lim_{n\to \infty}}$ ${|x_{n+1} - x|}$ = L ${( \lim_{n\to \infty}}$${|x_n - x|)}$.
which implies,
${ \lim_{n\to \infty}}$ ${|x_{n+1} - x|}$ ${\lt}$ ${( \lim_{n\to \infty}}$${|x_n - x|)}$.
Now how to find for any ${\epsilon \gt 0}$, there exists an M ${\in \mathbb{N}}$, such that
${|x_n - x| \lt \epsilon}$.
Please suggest me a method to prove the statement.