JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Limits XII

Put $t=x-1\to 0$. The numerator $-ax+\sin (x-1)+a$ becomes $\sin t-at$ and the denominator $x+\sin (x-1)-1$ becomes $t+\sin t$, so as $t\to 0$ the base tends to $\frac{t-at}{t+t}=\frac{1-a}{2}$. The exponent $\frac{1-x}{1-\sqrt{x}}=1+\sqrt{x}\to 2$. For the limit to be a genuine real number the base must stay positive as $x\to 1$, which requires $\frac{1-a}{2}>0$, i.e. $a<1$. Under that condition the limit is ${\left(\frac{1-a}{2}\right)}^2=\frac{1}{4}$, giving $\frac{1-a}{2}=\frac{1}{2}$ and $a=0$. The other algebraic root $a=2$ makes the base negative near $x=1$, where a real power is undefined, so it must be rejected, and the largest non-negative integer that works is $0$.