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Find Largest Integer values of $a$ for which $\displaystyle \lim_{x\rightarrow 1}\Bigg(\frac{-ax+\sin (x-1)+a}{x+\sin (x-1)-1}\Bigg)^{\frac{1-x}{1-\sqrt{x}}} = \frac{1}{4}\;,$ is

$\bf{My\; Solution::}$ We can write the above Limit as $\displaystyle \lim_{x\rightarrow 1}\Bigg(\frac{-a(x-1)+\sin(x-1)}{(x-1)+\sin(x-1)}\Bigg)^{1+\sqrt{x}} = \frac{1}{4}$

So $\displaystyle \lim_{x\rightarrow 1}\Bigg(\frac{-a+\frac{\sin(x-1)}{x-1}}{1+\frac{\sin (x-1)}{x-1}}\Bigg)^2 = \frac{1}{4}\Rightarrow \bigg(\frac{-a+1}{2}\bigg)^2 = \bigg(\frac{1}{2}\bigg)^2$

So We get $\displaystyle (-a+1) = \pm 1\Rightarrow a = 0$ or $a=2$

But answer given as $a = 0$ and I did not understand why it can not be $a = 2$

Please explain the reasoning behind it, Thanks

juantheron
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  • $[f(x)]^{g(x)}$ is defined when $f(x)>0$. When $a=2$, $f(x)=\frac{{ - 2\left( {x - 1} \right) + \sin \left( {x - 1} \right)}}{{x - 1 + \sin \left( {x - 1} \right)}} \leqslant 0$. – Baily Feb 18 '15 at 17:25

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