Find solution to multivariate equation in the infinite limit for one variable. I have the below equation (where $\{x|x\in\mathbb{R},0\lt x\lt \frac{1}{2}\}$ and $\{n|n\in\mathbb{N},n\geq5\}$):
$-(\frac{x}{1-x})^{n-1}=\frac{-x(n+2)+2}{-x(n+2)+n}$
A) Ideally, I want to determine a closed form solution for $x$ (in terms of $n$).
B) I want to determine the value of $x$ that satisfies the equation in the limit as $n\rightarrow\infty$.
I have the following results so far.
Bounds for the LHS:
$0\lt x\lt\frac{1}{2}\Rightarrow 0\lt\frac{x}{1-x}\lt 1\Rightarrow -1\lt-(\frac{x}{1-x})^{n-1}\lt 0$
Thus, the RHS satisfies:
$\frac{-x(n+2)+2}{-x(n+2)+n}\lt 0\Rightarrow -x(n+2)+2\lt 0\Rightarrow x\gt\frac{2}{n+2}$
My hypothesis is that $x\rightarrow\frac{2}{n+2}$ as $n\rightarrow\infty$. Empirically, $x\rightarrow\frac{2}{n+2}$ very fast (within $10^{-5}$ for $n$ as small as $10$).
But how do I prove this?
 A: A)
For the exact solution of $x$ in terms of $n$ for this $n^{th}$ degree polynomial equation, it seems it is always possible. At least WolframAlpha gives such closed forms for high $n$ as well (e.g. $n=20$: https://www.wolframalpha.com/input?i=-%28x%2F%281%E2%88%92x%29%29%5E%2820%E2%88%921%29%3D%28-x*%2820%2B2%29%2B2%29%2F%28-x*%2820%2B2%29%2B20%29).
How does WolframAlpha do this? How can I get the closed form myself?

B)
The following approach to try to show the limit seems to have a flaw, but maybe it's on the right track.
Since we know $x\gt\frac{2}{n+2}$, we can substitute $x=\frac{2}{n+2}+\epsilon$ into the equation (where $\epsilon>0$ can be arbitrarily small).
So the equation becomes
$-(\frac{\frac{2}{n+2}+\epsilon}{1-(\frac{2}{n+2}+\epsilon)})^{n-1}=\frac{-(\frac{2}{n+2}+\epsilon)(n+2)+2}{-(\frac{2}{n+2}+\epsilon)(n+2)+n}$
$-(\frac{2+\epsilon n+2\epsilon}{n-\epsilon n-2\epsilon})^{n-1}=\frac{-\epsilon n-2\epsilon}{n-2-\epsilon n-2\epsilon}$

We first evaluate the limit for the LHS:
$\lim\limits_{n\rightarrow\infty}{LHS}=\lim\limits_{n\rightarrow\infty}{-(\frac{2+\epsilon n+2\epsilon}{n-\epsilon n-2\epsilon})^{n-1}}$
Defining $f(n)=\frac{2+\epsilon n+2\epsilon}{n-\epsilon n-2\epsilon}$ and $g(n)=n-1$, we're trying to find $\lim\limits_{n\rightarrow\infty}{-f(n)^{g(n)}}=-\lim\limits_{n\rightarrow\infty}{\exp{(\ln{f(n)^{g(n)}})}}=-\lim\limits_{n\rightarrow\infty}{\exp{(g(n)\ln{f(n)})}}$.
Since they're continuous, this is $-\exp{(\lim\limits_{n\rightarrow\infty}{g(n)\ln{f(n)}})}=-\exp{(\lim\limits_{n\rightarrow\infty}{g(n)}\lim\limits_{n\rightarrow\infty}{\ln{f(n)}})}=-\exp{(\infty \ln{\frac{\epsilon}{1-\epsilon}})}$. Since $0\lt\epsilon\lt 1$, $\ln{\frac{\epsilon}{1-\epsilon}}<0$. So we end up with $-\exp{(-\infty)}=0$.
We now evaluate the limit for the RHS and set it to equal to $0$ (since it must match the LHS): $\lim\limits_{n\rightarrow\infty}{RHS}=\lim\limits_{n\rightarrow\infty}{\frac{-\epsilon n-2\epsilon}{n-2-\epsilon n -2\epsilon}}=\frac{-\epsilon}{1-\epsilon}=0$.
Thus, as $n\rightarrow\infty$, $\epsilon=0\Rightarrow x=\frac{2}{n+2}$.

The problem with this approach is that it gives $\epsilon=0$ upon substituting other values of $x$ as well (e.g. $x=\frac{3}{n+2}+\epsilon$ or $x=\frac{2}{n+2}+\frac{1}{n^2}+\epsilon$). Can I make the following argument somehow: We already know that $x\gt\frac{2}{n+2}$, so since the approach works upon substituting any value $x\gt\frac{2}{n+2}$, the limiting value must be $\frac{2}{n+2}$?
A: Answering your own answer

*

*Wolfram Alpha does not give you the exact solution : it just solve numerically for all roots of a polynomial of degree $20$. Just try this one in which I changed the $20$ to $20.1$. In what I wrote in my previous answer, you can treat $n$ as a real.


*If we follow your idea
$$x=\frac 2 {n+2}+\epsilon \quad \implies \quad  \left(\frac{2+(n+2) \epsilon }{n-(n+2) \epsilon }\right)^{n-1}+\frac{(n+2) \epsilon
   }{(2-n)+(n+2) \epsilon }=0$$ Expanding as Taylor series, we end with
$$0=2^{n-1} n^{1-n}+(n+2) \left(2^{n-2} (n-1) (n+2) n^{-n}+\frac{1}{2-n}\right)
   \epsilon +O\left(\epsilon ^2\right)$$ Ignoring the higher order terms
$$\epsilon=\frac{ n\,(n-2)} {n+2 }\,\,\frac{2^{(n+1)}}{4 n^n-2^n (n-2) (n-1) (n+2) }$$ where you see a kind of exponential trend
