Question

The largest value of the nonnegative integer $a$ for which
$\underset{x\to 1}{lim}{\left\{\frac{-ax+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4}$
is

Answer 1

$\underset{x\to 1}{lim}{\left\{\frac{-ax+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4}\Rightarrow \underset{x\to 1}{lim}{\left\{\frac{-ax+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}}^{\frac{(1-\sqrt{x})(1+\sqrt{x})}{1-\sqrt{x}}}=\frac{1}{4}\Rightarrow \underset{x\to 1}{lim}{\left\{\frac{-ax+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}}^{1+\sqrt{x}}=\frac{1}{4}\Rightarrow \underset{x\to 1}{lim}{\left\{\frac{\sin (x-1)-a(x-1)}{\sin (x-1)+x-1}\right\}}^{1+\sqrt{x}}=\frac{1}{4}\Rightarrow \underset{x\to 1}{lim}{\left\{\frac{\frac{\sin (x-1)}{(x-1)}-a}{\frac{\sin (x-1)}{(x-1)}+1}\right\}}^{1+\sqrt{x}}=\frac{1}{4}\Rightarrow {\left[\frac{1-a}{1+1}\right]}^2=\frac{1}{4}\Rightarrow \frac{1-a}{2}=\pm \frac{1}{2}\Rightarrow a=0,2$

Hence, $a=2$ is the largest non negative integer.