If $\alpha$ is the positive root of the equation, $p(x)=x^{2}-x-2=$
0 , then $\lim _{x \rightarrow \alpha^{+}} \frac{\sqrt{1-\cos (\mathrm{p}(x))}}{x+\alpha-4}$ is equal to:
Correct Option: , 2
$x^{2}-x-2=0 \Rightarrow(x-2)(x+1)=0$
$\Rightarrow x=2,-1 \Rightarrow \alpha=2$
$\therefore \lim _{x \rightarrow 2^{+}} \frac{\sqrt{1-\cos \left(x^{2}-x-2\right)}}{x-2}$
$=\lim _{x \rightarrow 2^{+}} \frac{\sqrt{2}\left|\sin \left(\frac{x^{2}-x-2}{2}\right)\right|}{x-2}$
$=\lim _{x \rightarrow 2^{+}} \frac{\sqrt{2} \sin \frac{\left(x^{2}-x-2\right)}{2}}{\left(\frac{x^{2}-x-2}{2}\right)} \times \frac{\left(x^{2}-x-2\right)}{2(x-2)}$
$=\frac{1}{\sqrt{2}} \lim _{x \rightarrow 2^{+}}\left(\frac{\sin \left(\frac{x^{2}-x-2}{2}\right)}{\frac{x^{2}-x-2}{2}}\right) \times \lim _{x \rightarrow 2^{+}} \frac{(x-2)(x+1)}{(x-2)}$
$=\frac{1}{\sqrt{2}} \times 1 \times 3=\frac{3}{\sqrt{2}}$
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