For $x_n>0$ for all $n\in\Bbb N$ prove that $\underline\lim\frac{x_{n+1}}{x_n}\le\underline\lim\sqrt[n]{x_n}\le\overline\lim \sqrt[n]{x_n}\le \overline\lim\frac{x_{n+1}}{x_n}$
HINT: if $q<\underline\lim\frac{x_{n+1}}{x_n}$ then $q\le\frac{x_{n+1}}{x_n}$ for all $n\ge N_q$.
- Here $\underline\lim$ mean limit inferior and $\overline\lim$ mean limit superior. Observe that the statement dont require that the sequence $(x_n)$ will be convergent.
I dont know how to prove this, some hint will be appreciated. My work so far: we can choose some $q>0$ that for some $N\in\Bbb N$ we have that
$$\left(q\le \frac{x_{n+1}}{x_n}\right)\land (q\le \sqrt[n]{x_n}),\quad\forall n\ge N_q$$
I dont get something useful from here (or at least I dont see that this is useful), an example
$$q^n\le x_n\le\frac{x_{n+1}}{q},\quad\forall n\ge N_q$$
I tried to get something by contradiction, i.e. that the statement $$\underline\lim\sqrt[n]{x_n}<\underline\lim\frac{x_{n+1}}{x_n}$$
is not possible. But again I dont found something enlightening. The unique thing that I worked is that if $(x_n)\to a$ is a constant sequence then the statement of the question is true.
Another crazy idea that I tried is choose some $0<q<1$, such that
$$\left(q\le\frac{x_{n+1}}{x_n}\right)\land (q\le x_n),\quad\forall n\ge N_q$$
then with little work we can see that
$$q\le\frac{x_{n+1}}{x_n}\implies q<\sqrt[m+n]{q^{m+1}}\le\sqrt[n+m]{x_{n+m}},\quad\forall n\ge N_q$$
but it is not clear that this last statement prove something useful. Well, any hint will be appreciated, thank you.
From the hint you have $ q \le x_{n+1} / x_n $ for each $ n \ge N_q $. Now you could use the telescopic product $ (x_{n+1}/x_n)...(x_{N_q+1}/x_{N_q})=x_{n+1}/x_{N_q} $.