Limit-The question asks to find the value of a in the limit. 
I did the whole sum,and my answer is a=0,2.
And,as the problem asks for the largest value is a=2.
However,the answer is given a=0.
I couldnot understand how it can be.Can any expain?
 A: It seems like you have correctly worked out the limits: 
$\displaystyle\lim_{x \to 1}\dfrac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1} = \dfrac{1-a}{2}$ and $\displaystyle\lim_{x \to 1}\dfrac{1-x}{1-\sqrt{x}} = \dfrac{1}{2}$
and then you solved the equation $\left(\dfrac{1-a}{2}\right)^{1/2} = \dfrac{1}{4}$ to get $a = 0,2$. 
However, you need to remember that $A^B$ is not well defined if $A$ is negative and $B$ is irrational. 
If $a > 1$, then $\dfrac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1}$ is negative for $x$ near $1$. 
Also, there are some values of $x$ arbitrarily close to $1$ such that $\dfrac{1-x}{1-\sqrt{x}} = 1+\sqrt{x}$ is irrational. 
Hence, $\left[\dfrac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1}\right]^{\frac{1-x}{1-\sqrt{x}}}$ is undefined for some values of $x$ arbitrarily close to $1$. 
Therefore, $\displaystyle\lim_{x \to 1}\left[\dfrac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1}\right]^{\frac{1-x}{1-\sqrt{x}}}$ does not exist for $a > 1$. 
A: Just to go in the same direction as JimmyK4542, around $x=1$, Taylor expansion is $$\dfrac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1}=\frac{1-a}{2}-\frac{a+1}{24}  (x-1)^2-\frac{a+1}{720}
   (x-1)^4+O\left((x-1)^5\right)$$ which is always negative for any $a>1$.
