A limit question (JEE $2014$)

For $a = 2$, we have - writing $x = 1 + \delta$ - the expression

$$\left(\frac{\sin\delta - 2\delta}{\sin\delta + \delta}\right)^{\Large\frac{\delta}{\sqrt{1+\delta}-1}}.\tag{1}$$

For most $0 < \lvert\delta\rvert < 1$, the exponent is not a rational number.

So the question is: Does raising a negative real number to a non-rational power make sense or not?

If you have no qualms about using complex numbers (like Wolfram), the expression makes sense, and the limit as $\delta\to 0$ is $\frac{1}{4}$, no problem.

The makers of the test apparently had qualms about using complex numbers and decided that the expression $(1)$ doesn't make sense in general.

Let's study the base of the exponential: $$ \lim_{x\to1}\frac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1}\overset{\mathrm{H}}{=} \lim_{t\to0}\frac{-a+\cos(x-1)}{1+\cos(x-1)}=\frac{1-a}{2} $$ (“H” denotes an application of l'Hôpital's theorem). Thus the limit is non negative only for $a\le1$. In particular, being the base continuous after extending it to $0$ with the limit, if $a>1$ there is a neighborhood of $1$ where your function is not defined (the basis of an exponential must be non negative).

The limit now poses no problem, because the exponent is $1+\sqrt{x}$ (for $x\ne0$), so your limit is $$ \left(\frac{1-a}{2}\right)^2 $$ which is equal to $1/4$ if and only if $$ \left(\frac{1-a}{2}=\frac{1}{2} \quad\text{or}\quad \frac{1-a}{2}=-\frac{1}{2}\right) \quad\text{and}\quad a\le1 $$ that is, $a=0$.

Whatever the (finite) limit inside the main parenthesis ($\frac{1-a}2$), the limit of the exponent is $2$ (it simplifies to $x+\sqrt x$), and the full expression is positive.

Solution: $a=2.$

UPDATE:

The power of a negative (when $a=2$) can be dealt with with complex logarithms: $$\lim_{x\rightarrow1}\ln(-|f(x)|)^{g(x)}=\lim_{x\rightarrow1}(g(x)(\ln |f(x)|+i\pi))=\lim_{x\rightarrow1} g(x)\lim_{x\rightarrow1}(\ln |f(x)|+i\pi)=2\ln\frac12+2i\pi.$$