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

The largest value of the non negative integer ‘a’ for which
${lim}_{x\to 1}{\left\{\frac{-ax+sin(x-1)+a}{x+sin(x-1)-1}\right\}}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4}$
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
${lim}_{x\to 1}f\left(x\right)={lim}_{x\to 1}\frac{cos(x-1)-a}{cos(x-1)+1}=\frac{cos0-a}{cos0+1}=\frac{1-a}{2}$
$g\left(x\right)=\frac{1-x}{1-\sqrt{x}}$
${lim}_{x\to 1}g\left(x\right)={lim}_{x\to 1}\frac{1-x}{1-\sqrt{x}}=\frac{0}{0}$
Using L’hospital’s theorem
${lim}_{x\to 1}g\left(x\right)={lim}_{x\to 1}\frac{-1}{\frac{-1}{2\sqrt{x}}}=1\times 2=2$
${lim}_{x\to 1}\left[f\right(x){]}^{g\left(x\right)}=\frac{1}{4}ln{lim}_{x\to 1}[f\left(x\right){]}^{g\left(x\right)}=ln\frac{1}{4}{lim}_{x\to 1}g\left(x\right)lnf\left(x\right)=ln\frac{1}{4}{lim}_{x\to 1}g\left(x\right).{lim}_{x\to 1}lm\left(f\right(x\left)\right)=ln\frac{1}{4}2.ln{lim}_{x\to 1}f\left(x\right)=ln\frac{1}{4}2.ln\left(\frac{1-a}{2}\right)=ln\frac{1}{4}{\left(\frac{1-a}{2}\right)}^2=\frac{1}{4}$
Solving this we get $a=0$ and $a=2$.
But for $a=2$ we get a negative number raised to the power of a rational number. This is not always defined.
Hence $a=0$ is the right answer.