Is $|x-2x_k|<\varepsilon$ for all $k>N$ sufficient to show a sequence is convergent?

Question

I am about to start with a few mathematics courses next semester and I was hoping someone could help me out with the following question.

Suppose there is a real-valued sequence $(x_k)$ such that there is a number $x\in\mathbf{R}$ such that for any $\varepsilon>0$ there is a positive integer $N$ such that $|x-2x_k|<\varepsilon$ for all $k>N$.

My gut feeling says that this sequence is convergent, since the $x_k$ will become arbitrary close to $x$. Is there a neat way to write this down?

Answer 1

At first sight, I would say this is exactly the definition of $\frac{x}{2}$ as the limit, so the answer is yes.

Answer 2

Yes, the sequence $(x_{k})$ converges to $x/2$. To see this, note that for every $\varepsilon > 0$ there is some $N \geq 1$ such that $|\frac{x}{2} - x_{k}| = \frac{1}{2}|x-2x_{k}| < \varepsilon$ for all $k \geq N$.