Is it possible for the sequence $\{\frac{x_{n+1}}{x_n}\}$ to be unbounded but have $\lim_{n\to\infty} x_n = x$, $x_n \ne 0$

Given: $$\begin{cases} \lim_{n\to\infty} x_n = x \\ x_n \ne 0 \\ n \in \mathbb N \end{cases}$$ Is it possible for $$\{\frac{x_{n+1}}{x_n}\}$$ to be and unbounded sequence?

This problem comes in the context of two others which are:

1. Does $$\lim_{n\to\infty}\frac{x_{n+1}}{x_n}$$ exist?

Not necessarily if we for example consider: $$x_n = \frac{sin{\pi n \over \sqrt2}}{n}$$

1. If the limit exists and is equal to $$q$$, prove $$|q| \le 1$$

This one is also easy to show using the definition of a monotone sequence.

The third part as of the question section asks whether $$\{\frac{x_{n+1}}{x_n}\}$$ may be unbounded, and the answer suggests that this is indeed possible, but I don't see how.

What would be such a sequence?

Take $$x_n = \begin{cases} \dfrac{1}{n}, & \text{n odd} \\[6pt] \dfrac{1}{n^2}, & \text{n even.} \end{cases}$$
(If we assume $$x \neq 0$$, there is no such example.)