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.

Use that $|a+b|\ge ||a|-|b||$, so that
$$
\lim_{n\to \infty} \left||x_{n+1}|-\left|\frac{x_n}{2}\right|\right|=0.
$$
For each $\varepsilon>0$ there exists $n_0$ such that if $n\ge n_0$ then
$$
\left||x_{n+1}|-\left|\frac{x_n}{2}\right|\right|\le \varepsilon \, \implies \, |x_{n+1}| \le \left|\frac{x_n}{2}\right|+\varepsilon
$$
Assume wlog $n_0=0$, since we do not care a finite number of terms. Then by induction
$$
|x_{n}| \le \left|\frac{x_0}{2^n}\right|+2\varepsilon.
$$
The claim follows by the arbitrariness of $\varepsilon$.
