Showing $|x_n -x|$ converges to x I'm given a generic sequence $x_n$, and I know that $\lim_{n ->\infty} \frac{|x_{n+1}-x|}{|x_n-x|} < 1$ and exists. I'm trying to show that $x_n$ converges to $x$. 
I tried a bunch of approaches, but the farthest I got was a proof by contradiction where I try show that $\lim_{n->\infty} |x_n-x|$ cannot be greater than 0, but instead must equal 0, resulting in $x_n$ converging to $x$. 
Can I get some assistance?

Hint:
It is equivalent to show tha $|x_n-x|\to 0$. Now, since $\lim_{n\to\infty} \frac{|x_{n+1}-x|}{|x_n-x|} $ exists and is $<1$, there exists $k<1$ and $N_0$ such that
$$\frac{|x_{n+1}-x|}{|x_n-x|}\le k\quad\forall n \ge N_0.$$
Deduce that, for all $n\ge N_0$, one has $\;|x_n-x|\le k^{n-N_0}\,|x_{N_0}-x|$ (use induction on $n$) and conclude.
