What is the limit of this sequencen pOokiqकu Zpretसाेष्9Acl
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 ?
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$\\begingroup$ $x_n >1$ and $\\lim x_n \\geq 1$ are both unnecessary. See my answer. $\\endgroup$ – Kavi Rama Murthy 6 hours ago
3 Answers
$\\lim_{n\\rightarrow \\infty}x_n = 1$. 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 $\\lim x_n=1$.
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 $\\lim x_n=1$.
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 $x_n \\to 1$.
