Super tricky limit problem

We have $\sin (x-1) =(x-1) -\frac{(x-1)^3}{3!} + \frac{(x-1)^5}{5!} +\cdots$ $\displaystyle\lim_{x\to1}\left\{\dfrac{-ax + \sin(x-1) + a}{x + \sin(x-1) - 1}\right\}^\frac{1-x}{1-\sqrt{x}} = \dfrac{1}{4}\\ \displaystyle\lim_{x\to1}\left\{\dfrac{-a(x-1)+\sin(x-1) }{x -1 +\sin(x-1) }\right\}^{1+\sqrt x} = \dfrac{1}{4} \\ \displaystyle\lim_{x\to1}\left\{\dfrac{-a(x-1)+\left(x-1 -\frac{(x-1)^3}{3!} +\frac{(x-1)^7}{7!}+\cdots \right) }{x -1 +\left(x-1 -\frac{(x-1)^3}{3!} +\frac{(x-1)^7}{7!}+\cdots \right) }\right\}^{1+\sqrt x} = \dfrac{1}{4} \\ \displaystyle\lim_{x\to1}\left\{\dfrac{-a +\left(1 -\frac{(x-1)^3}{3!(x-1)} +\frac{(x-1)^7}{7!(x-1)}+ \cdots \right) }{1 +\left(1 -\frac{(x-1)^3}{3!(x-1)} +\frac{(x-1)^7}{7!(x-1)}+\cdots \right) }\right\}^{1+\sqrt x} = \dfrac{1}{4} \\ \displaystyle\lim_{x\to1}\left\{\dfrac{-a +\left(1 -\frac{(x-1)^2}{3!} +\frac{(x-1)^6}{7!}+ \cdots \right) }{1 +\left(1 -\frac{(x-1)^2}{3!} +\frac{(x-1)^6}{7!}+\cdots \right) }\right\}^{1+\sqrt x} = \dfrac{1}{4} \\ \displaystyle\lim_{x\to1}\left\{\dfrac{-a +1 }{1 +1 }\right\}^{1+\sqrt 1} = \dfrac{1}{4} \implies \dfrac{(-a+1)^2}{4}=\dfrac{1}{4} \implies a = 0, 2$ The answer that satisfy the limit is $0$ .

In response to Fish Bacon query . If we input $a =2$ then limit will be undefined since function we have is like $f(x)^{g(x)}$ and as $x\to 1$ , $g(x)$ is approching $2$ and $f(x) <0$ which undergoes the complex form. To be limit defined $f(x)\geq 0$ which is only possible if $a=0$ .