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 August 7th, 2011, 10:39 AM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 limit with polynomial If $x_{n}$ be the positive root of the equation $x^n= x^2+x+1$ Then find $\lim_{n\rightarrow \infty}n.\left(x_{n} - 1\right)=$
 August 7th, 2011, 01:52 PM #2 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: limit with polynomial $x_{n}^{2}+x_{n}+1=x^{n}$ If $x_{n}=1$, then we get $x^{n}=3$. Thus, $x_{n}^{n}>3$ $x_{n}>3^{\frac{1}{n}}$ So, evaluate the limit $\lim_{n\to \infty}n(3^{\frac{1}{n}}-1)$
 August 7th, 2011, 06:14 PM #3 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 Re: limit with polynomial thanks galactus but can anyone explain me first line. $x^2_{n}+x_{n}+1= x_{n}$
 August 9th, 2011, 05:36 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: limit with polynomial You have miscopied. The first line was $x_n^2+ x_n+ 1= x_n^n$. The right side is "$x_n^n$", not just "$x_n$". You said, in your first post, that $x_n$ is a root of $x^n= x^2+ x+ 1$. That means that $x_n$ satisfies $x_n^2+ x_n+ 1= x_n^n$.

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