If $\lim_{n\to\infty} x_{n+1}-\frac12 x_n = 0$ then $x_n\to 0$ 
Let $x_n$ be a sequence of real numbers such that $\displaystyle\lim_{n\to\infty} x_{n+1}-\frac12 x_n = 0$
Prove that $\lim_{n\to\infty} x_n = 0$

I have a proof if one assumes that $(x_n)$ is bounded. In that case, $(x_n)$ has as accumulation point, say $\alpha$, and the hypothesis $\lim_{n\to\infty} x_{n+1}-\frac12 x_n = 0$ implies that $2\alpha$ is also an accumulation point of $(x_n)$.
Hence $\forall p \;, \; 2^p\alpha$ is an accumulation point, which contradicts the boundedness assumption, unless $\alpha =0$. $0$ is  consequently the only accumulation point of $(x_n)$ and the claim is proved.
It remains to show that $(x_n)$ is bounded. I've tried contradiction for that matter, without success.

Alternatively, rewrite the hypothesis as $$\lim_{n\to\infty}x_n\left(\frac{x_{n+1}}{x_n}-\frac{1}{2}\right)=0.$$ Even if we can't split the limit, as we don't know a priori that $x_n$ converges, we do know we must have either $x_n\to0$ or $x_{n+1}/ x_n \to 1/2 $, which still implies $x_n\to0$.
