Limit of non-negative real root of $x^n+x^{n-1}+x-1=0$ as $n \to \infty$ Let $x_n$ be the unique non-negative real solution of a equation $x^n+x^{n-1}+x-1=0$. Prove the sequence {$x_n$} is increasing and converges to 1.
I found out that $ 0 < x_n < 1$ but just that..
 A: Let us define $f_n(x)=x^n+x^{n-1}+x-1$.  For all $n, f_n(0)=-1, f(1)=2$ so there is a root in $(0,1)$.  We are told there is only one positive root, so this is it.  Intuitively, as $n$ gets larger the terms $x^n, x^{n-1}$ will get smaller so $x$ has to grow to compensate.  Note that for all $n, f_n(x)$ is increasing on $(0,1)$.  We are given that $f_n(x_n)=0$.  Now evaluate $f_{n+1}(x_n)=x_n^{n+1}+x_n^n+x_n-1=x_n^{n+1}-x_n^{n-1}\lt 0$ so the root of $f_{n+1}$, which is $x_{n+1}$ is in $(x_n,1)$ and is greater than $x_n$.  By the way, the sequence is bounded above by $1$ and increasing, so is convergent, so is Cauchy.
A: Let us call the unique non-negative real solution of $f_n(x) = x^n + x^{n-1} + x - 1 = 0$ as $y_{n}$. 
By the intermediate value theorem, $f_n(0) = -1$ and $f_n(1) = 2$, so $0 < y_n < 1$ is true.
Note that for $0 < x <1$ and $n < m$, we have $x^m < x^n$, and therefore $x^{m+1} + x^m + x  - 1 < x^{n+1} + x^n + x - 1$. That is, on this interval, $f_n(x) > f_m(x)$.
Therefore, note that $f_n(y_m) > f_m(y_m) > 0$, so by the intermediate value theorem, $y_n \in (0,y_m)$, so that $y_n < y_m$. Therefore, $y_n$ is an increasing sequence.
It is bounded, and therefore must have a limit. But why is this limit equal to $1$? It might as well be anything between $0$ and $1$. This would stem from an argument of this kind:

Let $x_0 < 1$. We will prove that for some large enough $m$, $y_m > x_0$. To see this, note that we can rewrite $f_n(x_0)$ as $x_0(x_0^{n-1} + x_0^n + 1) - 1$. By sufficiently increasing $n$, and noting that $x_0^n \to 0$ as $n \to \infty$, we see that $(x_0^{n-1} + x_0^n + 1)$ can be brought below $\frac 1{x_0}$, since the former has limit $1$, while the latter is some quantity greater than $1$. That is, for some $N$, we have $f_N(x_0) < 0$. By the intermediate value theorem, again, we conclude that $y_N > x_0$ for this $N$.

