How prove this limit $\displaystyle\lim_{n\to \infty}a_{n}=1$ Let $a_n$ be the only positive root of the equation $x^n+x=1$, for each $n\in \Bbb N$.
Show that $\lim \limits_{n\to \infty}a_{n}$ exists,and find its value.
My guess is that
$$\lim \limits_{n\to \infty}a_{n}=1$$
But How prove that it exists, and how find it?
 A: You may show the sequence of roots is increasing, and bounded above by $1$, say. Rebecca's approach then shows the limit of the roots cannot be $<1$, so it must be $\geqslant 1$. Having shown it is $\leqslant 1$; we get it must be $1$.
If you want to be sketchy, consider the following. The functions $f_n=x^n$ are always increasing for $x>0$, with $f_n(0)=0$ and $f_n(1)=1$. The function $f(x)=1-x$ is monotone decreasing and $f(0)=1$; $f(1)=0$. You can show the $f_n=f$ has a root, and it must be $<1$. As $n$ grows, $f_n$ decreases over $0<x<1$ (so the ordinate of the root increases); with pointwise limit $=0$, so $1-x=0\implies x=1$.
A: Technically, one should first verify that $a_n$ is indeed unique.
This can be done easiest by Descartes' Rule of Signs, a less-known but in my opinion underrated result which is used as follows.  Note that the polynomial $x^n + x - 1$ has coefficients $1, 1, -1$ in that order (ignore $0$ coefficients), so the coefficients signs go $+,+, -$ (positive, positive, negative).  The coefficients change sign exactly once, so the number of positive real roots of the polynomial $x^n + x - 1$ is one.  (In general, if the coefficients change sign $n$ times, the number of positive real roots will be one of $n$, $n - 2$, $n - 4$, $n - 6$, etc.)
Once you have established that $a_n$ is unique, Rebecca's solution completes the proof.
A: It is sufficient to show that, for all $0<\varepsilon<1$, the function $f$, defined by $f_n(x)=x^n+x-1$, has a root in the interval $(1-\varepsilon,1)$ for all sufficiently large $n$.
We know $f_n(1)=1>0$.  Thus, since $f_n$ is continuous, it sufficient to show that $f(1-\varepsilon)<0$ for all sufficiently large $n$.
We observe
\begin{align*}
f_n(1-\varepsilon) &= (1-\varepsilon)^n+1-\varepsilon-1 \\
 &=(1-\varepsilon)^n-\varepsilon \\
 & \rightarrow -\varepsilon \\
 & < 0
\end{align*}
since $c^n \rightarrow 0$ whenever $0 < c < 1$.
Hence, for all $0<\varepsilon<1$, we have shown that $f_n(1-\varepsilon)<0$ for all sufficiently large $n$.
