The following is a JEE (A national level entrance test) question:
Find the largest value of the non-negative integer ( a ) for which:
$$ \displaystyle \lim_{x \to 1} \left( \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } \right)^{\dfrac{1-x}{1-\sqrt x} } = \dfrac 1 4 $$
On solving this, we get $ a = 0 $ or $ a = 2 $. So we take the answer as $ 2 $. But the official answer key says that the answer is $ 0 $. The reason given (not in the official key – they simple contain the answer. I saw the reason in an "unofficial" solution to the question paper) is that if $ a = 2 $, then the term $\displaystyle \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } $ tends to a negative value ( $-0.5$ ). So $ a = 0$.
BUT:
By putting $a = 2$ in the limit, and inputting that to wolframalpha (http://www.wolframalpha.com/input/?i=lim(+(+(-2x+%2B+2+%2B+sin(x-1)+)+%2F+(+x+-+1+%2B+sin(+x+-+1)+)+)+%5E+(+(1-x)%2F(1-sqrt(x))+)+)+as+x+tends+to+1), the answer turns out to be $0.25$. So, by putting $ a = 2$ also, we get the same value of the limit – $0.25$.
How to solve this dispute? Also, could you please check with Mathematica? Unfortunately, I don't have access to it!
(This is very important as the change in the answer key will affect the rankings in a dramatic way)
Best Answer
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.