Question: Consider the following fixed-point equation:

\begin{equation}
x = (1-x)^{1-\frac{2}{a+1}} - 1,
\end{equation}
where $x \in [0,1]$ and $a \in [0,1]$. If we write the solution of $x$ in terms of $a$ as $x(a)$, then, would it be possible to compute $\lim\_{a \rightarrow 1} x(a)$ ?

Since there is no closed-form solution for $x$, I tried to numerically see what happens. Here is the result:

$a=0.5640 \Leftrightarrow x=0.9$;

$a=0.7399 \Leftrightarrow x=0.99$;

... ...

$a=0.9547 \Leftrightarrow x=0.9999999999999$.

But is there any analytic way to demonstrate $\lim\_{a \rightarrow 1} x(a)$ ?

Answer: 
Moving -1 to the other side and taking logs yields
$$
\begin{split}
\ln(x+1) &= \left(1-\frac{2}{a+1}\right)\ln(1-x) \\
\frac{\ln(x+1)}{\ln(1-x)} &= 1-\frac{2}{a+1} \to 0 \text{ as } a \to 1
\end{split}
$$
so you are seeking the value of $x$ that will make the left-hand side $0$.

One options is to send the numerator to $0$, so $\ln(x+1) \to 0$ and thus $x+1 \to 1 \leftrightarrow x \to 0$ and so the LHS becomes 0/0, using L'Hospital's rule we can show that LHS is then $\frac{x-1}{x+1} \to -1$ so this way does not work.

Another way is to send the denominator to $\pm \infty$. Consider $-\infty$ first: so $\ln(1-x) \to -\infty$ and $1-x \to 0$ and $x \to 1$.

Finally, to $+\infty$: so $\ln(1-x) \to +\infty$ so $1-x \to +\infty$ and $x \to -\infty$, but that would make the numerator undefined.

In summary, the only choice is indeed $x \to 1$, which sends $\ln(1-x) \to -infty$ and the equality takes the form 0=0.
If you isolate $a$ in terms of $X$, the only possible solution as $\lim\_{a\to 1} x(a)$ appears to be $x=0$...