# Prove that $\lim \inf \dfrac{x_{n+1}}{x_n}\leq \lim \inf x_n^{1\over n} \leq \lim \sup x_n^{1\over n} \leq \lim \sup \dfrac{x_{n+1}}{x_n}.$

If $(x_n)$ is any sequence of positive real numbers, then $$\lim \inf \dfrac{x_{n+1}}{x_n}\leq \lim \inf x_n^{1\over n} \leq \lim \sup x_n^{1\over n} \leq \lim \sup \dfrac{x_{n+1}}{x_n}.$$

I don't understand how am I suppose to use the hint nor what to do afterwards.

Hint: If $\lim \sup \dfrac{x_{n+1}}{x_n}-\infty$, right hand inequality is obvious. So suppose $\lim \sup \dfrac{x_{n+1}}{x_n}=M$. Then $\dfrac{x_{n+1}}{x_n}< M+\epsilon$ $\forall n\geq N$, i.e., $x_{n+k}\leq (M+\epsilon)^kx_n$ for all $k\geq 0$.

If $$\limsup\frac{x_{n+1}}{x_n}=M<\infty,$$ then for every $\varepsilon>0$, there exists an $N\in\mathbb N$, such that $$\frac{x_{n+1}}{x_n}\le M+\varepsilon, \quad n\ge N.$$ Thus $$\frac{x_n}{x_N}=\frac{x_{N+1}}{x_N}\cdots\frac{x_{n}}{x_{n-1}}\le (M+\varepsilon)^{n-N},$$ and hence $$x_n\le \frac{x_N}{(M+\varepsilon)^N}\cdot (M+\varepsilon)^n\quad\Rightarrow\quad x_n^{1/n}\le \left(\frac{x_N}{(M+\varepsilon)^N}\right)^{1/n}\cdot (M+\varepsilon) \to M+\varepsilon.$$ Therefore $$\limsup x_n^{1/n}\le M+\varepsilon,\quad \text{for all \varepsilon>0.}$$