Evaluate the limit:$$\lim_{x\to 1}\left(\frac{2-2x+\sin(x-1)}{x-1+\sin(x-1)}\right)^{\frac{1-x}{1-\sqrt x}}$$
The answer given is $\frac{1}{4}$.
My Attempt
$$\lim_{x\to 1}\left(\frac{-2+\frac{\sin(x-1)}{x-1}}{1+\frac{\sin(x-1)}{x-1}}\right)^{1+\sqrt x}=\left(\frac{-2+1}{1+1}\right)^{1+1}=\frac{1}{4}$$
Let $f(x)=\left(\frac{-2+\frac{\sin(x-1)}{x-1}}{1+\frac{\sin(x-1)}{x-1}}\right)^{1+\sqrt x}$
Now though I do get the answer but as $x\to 1$ the base of the limit is a negative quantity. So limit should not exist. But at the same time I feel that $x$ is required to approach $1$ through values which lie in the domain of the function so the values at which $f(x)$ does not exist do not count so there is no problem with answer being $\frac{1}{4}$
Best Answer
This kind of issue has been already discussed in a different case here
In the present case, with reference to the more general definition of limit, we can give a meaning to the expression considering rational exponents with odd denominator reduced to coprime factors as discuss in this other answer.
In this way we can define a deleted neighborhood at the point $x=1$ and take the limit.