The largest value of the non-negative integer | vedclass

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(A) Let $L = \lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1-\sqrt{x}}}$.Since $\frac{1-x}{1-\sqrt{x}} =

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(A) Let $L = \lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1-\sqrt{x}}}$.Since $\frac{1-x}{1-\sqrt{x}} = \frac{(1-\sqrt{x})(1+\sqrt{x})}{1-\sqrt{x}} = 1+\sqrt{x}$,the limit becomes $\lim _{x \rightarrow 1}\left\{\frac{-a(x-1)+\sin(x-1)}{(x-1)+\sin(x-1)}\right\}^{1+\sqrt{x}} = \frac{1}{4}$.Let $x-1 = h$. As $x \rightarrow 1$,$h \rightarrow 0$. The expression becomes $\lim _{h \rightarrow 0}\left\{\frac{-ah+\sin h}{h+\sin h}\right\}^{1+\sqrt{1+h}} = \frac{1}{4}$.Dividing numerator and denominator by $h$,we get $\lim _{h \rightarrow 0}\left\{\frac{-a+\frac{\sin h}{h}}{1+\frac{\sin h}{h}}\right\}^{1+\sqrt{1+h}} = \frac{1}{4}$.Substituting $h=0$,we get $\left(\frac{-a+1}{1+1}\right)^{1+1} = \frac{1}{4}$,which simplifies to $\left(\frac{1-a}{2}\right)^2 = \frac{1}{4}$.This implies $\frac{1-a}{2} = \frac{1}{2}$ or $\frac{1-a}{2} = -\frac{1}{2}$.If $\frac{1-a}{2} = \frac{1}{2}$,then $1-a = 1$,so $a=0$.If $\frac{1-a}{2} = -\frac{1}{2}$,then $1-a = -1$,so $a=2$.For $a=2$,the base $\frac{-2h+\sin h}{h+\sin h} \approx \frac{-h}{2h} = -\frac{1}{2}$ as $h \rightarrow 0$. Since the base is negative,the power $1+\sqrt{1+h} \approx 2$ makes the result positive,but the limit of the base itself is negative,which is generally undefined in real-valued limits of the form $f(x)^{g(x)}$. Thus,$a=0$ is the valid non-negative integer.

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

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