Inferior and superior limit inequality

For any sequence $ x_n\in R, x_n>0 $ we have

$$\varliminf_{n \to +\infty} \frac{x_{n+1}}{x_{n}}\leq\varliminf_{n \to +\infty} \sqrt[n]{ x_{n}}\leq\varlimsup_{n \to+\infty}\sqrt[n]{ x_{n}}\leq\varlimsup_{n \to +\infty}\frac {x_{n+1}}{ x_{n}}$$

From here we also have
Proposition
For any $ (x_n)_{n\in N}, x_n>0 $ if

$$\lim_{n \to +\infty}\frac{x_{n+1}}{x_n}=l$$

then

$$\lim_{n \to +\infty}\sqrt[n]{ x_{n}}=l$$

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