As an alternative:$\displaystyle x_{n+1}-\tfrac12x_n=0 \implies x_{n+1}-x_n=-\tfrac12x_n$, so the sequence $\displaystyle \{x_n\}$ is decreasing when $\displaystyle x_n$ is positive and increasing when $\displaystyle x_n$ is negative.

Also $\displaystyle x_{n+1}=\tfrac12x_n$ tells us that $\displaystyle x_{n+1}$ has the same sign as $\displaystyle x_n$.

Thus, the sequence $\displaystyle \{x_n\}$ is either increasing and bounded above (by zero), or it is decreasing and bounded below (by zero), or (trivially) every $\displaystyle x_n$ is equal to zero. This tells us that the series converges.

When a series converges, $\displaystyle \lim_{n \to \infty} x_{n+1}-x_{n} =0$. But since $\displaystyle x_{n+1}=\tfrac12x_n$, we have $\displaystyle \lim_{n \to \infty} \tfrac12x_n-x_{n} =\lim_{n \to \infty} -\tfrac12x_n =0$ and so $\displaystyle \lim_{n \to \infty} x_{n} =0$ as required.