Question

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

Answer 1

Given :
${lim}_{x\to 1}{\left\{\frac{-ax+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4}$

It is in the form of ${lim}_{x\to a}\left[f\right(x){]}^{g\left(x\right)}=k$

Let $f\left(x\right)=\frac{-ax+\sin (x-1)+a}{x+\sin (x-1)-1}$

$f\left(x\right)=\frac{\sin (x-1)-0-a(x-1)}{\sin (x-1)+(x-1)}$

${lim}_{x\to 1}f\left(x\right)=\frac{0}{0}$

Using L’hospital’s theorem

${lim}_{x\to 1}f\left(x\right)={lim}_{x\to 1}\frac{\cos (x-1)-a}{\cos (x-1)+1}=\frac{\cos (1-1)-a}{\cos (1-1)+1}=\frac{\cos 0-a}{\cos 0+1}=\frac{1-a}{2}$

$\therefore {lim}_{x\to 1}f\left(x\right)=\frac{1-a}{2}$---(1)

Now, $g\left(x\right)=\frac{1-x}{1-\sqrt{x}}$

${lim}_{x\to 1}g\left(x\right)={lim}_{x\to 1}\frac{1-x}{1-\sqrt{x}}=\frac{1-1}{1-1}=\frac{0}{0}$
Using L’hospital’s theorem
${lim}_{x\to 1}g\left(x\right)={lim}_{x\to 1}\frac{-1}{\frac{-1}{2\sqrt{x}}}=1\times 2=2$

$\therefore {lim}_{x\to 1}g\left(x\right)=2$---(2)

Now,

${lim}_{x\to 1}\left[f\right(x){]}^{g\left(x\right)}=\frac{1}{4}$

apply $\ln$ on both sides,

$\ln {lim}_{x\to 1}\left[f\right(x){]}^{g\left(x\right)}=\ln \frac{1}{4}$

$\ln {lim}_{x\to 1}g\left(x\right)\ln f\left(x\right)=\ln \frac{1}{4}$

$\ln {lim}_{x\to 1}g\left(x\right).{lim}_{x\to 1}\ln \left(f\right(x\left)\right)=\ln \frac{1}{4}$

$2.\ln {lim}_{x\to 1}f\left(x\right)=\ln \frac{1}{4}$

$2.\ln \left(\frac{1-a}{2}\right)=\ln \frac{1}{4}$ --(3)

$\Rightarrow {\left(\frac{1-a}{2}\right)}^2=\frac{1}{4}$

$\Rightarrow \left(\frac{1-a}{2}\right)=\pm \frac{1}{2}$

$\Rightarrow \left(\frac{1-a}{2}\right)=\frac{1}{2}$ and $\left(\frac{1-a}{2}\right)=-\frac{1}{2}$

$\Rightarrow 1-a=1$ and $1-a=-1$

$\Rightarrow a=0$ and $a=2$.
But for $a=2$ we get a negative number raised to the power of a rational number.

and logarithm of a negative number is not define. [From (3)]

This is not always defined.
Hence $a=0$ is the right answer.