When is the product of limsups equal to the limsup of the products?

Question

Let $\left\{x_{n}\right\}$ be a sequence of real numbers where $x_{n} > 0$ for all $n \in \mathbb{N}$. Given that \begin{equation*} x = \limsup_{n \rightarrow \infty} x_{n}^{1/n} = \limsup_{n \rightarrow \infty} \frac{1}{x_{n}^{1/n}} \end{equation*} I am trying to show that $x = 1$. My reasoning so far is that \begin{align} x^{2} &= \left(\limsup_{n \rightarrow \infty} x_{n}^{1/n}\right) \left(\limsup_{n \rightarrow \infty} \frac{1}{x_{n}^{1/n}}\right)\\ &= \limsup_{n \rightarrow \infty} \left(x_{n}^{1/n}\right) \left(\frac{1}{x_{n}^{1/n}}\right)\\ &= \limsup_{n \rightarrow \infty} 1 = 1 \end{align} and so \begin{equation*} \sqrt{x^{2}} = \sqrt{1} \Rightarrow x = 1 \end{equation*} However is it true in this particular case that the product of the limsups is equal to the limsup of the product?

Answer 1

No take sequences $\frac 12,2,\frac 12,2,\frac 12,2,\cdots$ and the same shifted sequence $2,\frac 12,2,\frac 12,2,\cdots$

The term to term product is always $1$ while the product of $\varlimsup$ is $4$.

Although here it is a bit different because you do not have two unrelated sequences but the same $x_n$ so one limsup would be $2$ while the other would be $\frac 12$ clearly not equal.

Maybe considering $\varliminf x_n^{\frac 1n}$ instead can help.

Then if $\varlimsup$ and $\varliminf$ are equal you can conclude.