What is the limit of this sequence

Question

If $x_n$ is a sequence of real numbers greater than 1 and $\lim_{n \to \infty} x_n \geq 1$. Can we determine the limit of $x_n$ if we know that $\lim_{n \to \infty} x_n^n = 1$ ? If not, what conditions can we add to be able to determine the limit ?

Answer 1

$\lim_{n\rightarrow \infty}x_n = 1$. Since $x_n\geq1,\forall n\geq 1$, we have $x^n_n\geq x_n$, hence $\lim x_n^n\geq \lim x_n \geq 1$, which implies $\lim x_n=1$.

Answer 2

There is too much of unnecessary hypothesis in this question. Let $\{x_n\}$ be any sequence of real numbers such that $x_n^{n} \to 1$. Then $x_n >0$ after some stage . Taking logarithm we get $nlog \, x_n \to 0$. Since $\log x_n =\frac 1 n log x_n$ we get $\lim \log \, x_n=0$ or $\lim x_n=1$.

Answer 3

As $\lim_{n \to \infty} x_n^n = 1$, we have $x_n^n < 2$ for large enough $n$, so $x_n < \sqrt[n]{2}$. Also $x_n \geqslant 1$. As $\sqrt[n]{2} \to 1$ and $1 \to 1$, by squeeze theorem we have $x_n \to 1$.