If α is the positive root of the equation, p(x) = x ^ 2 - x - 2 = 0...

Given, $p\left(x\right) = 0$

$⇒ {x}^{2} - x - 2 = 0$

$⇒ \left(x - 2\right)\left(x + 1\right) = 0$

$⇒ x = 2, - 1 ∴ \alpha = 2$

Now,$\lim_{x \to {2}^{+}}\dfrac{\sqrt{1 - \cos \left({x}^{2} - x - 2\right)}}{x - 2}$

$= \lim_{x \to {2}^{+}}\dfrac{\sqrt{2\sin^{2}\left(\dfrac{{x}^{2} - x - 2}{2}\right)}}{x - 2}$

$= \lim_{x \to {2}^{+}}\sqrt{2} \cdot \dfrac{\left|\sin\left(\dfrac{{x}^{2} - x - 2}{2}\right)\right|}{x - 2}$

$= \lim_{x \to {2}^{+}}\sqrt{2} \cdot \dfrac{\sin \left(\dfrac{{x}^{2} - x - 2}{2}\right)}{x - 2}$ (because $x > 2$)

$= \lim_{x \to {2}^{+}}\dfrac{\sqrt{2}\sin \left(\dfrac{{x}^{2} - x - 2}{2}\right)}{\left(\dfrac{{x}^{2} - x - 2}{2}\right) \cdot 2} \cdot \left(\dfrac{{x}^{2} - x - 2}{x - 2}\right)$

$= \lim_{x \to {2}^{+}}\dfrac{1}{\sqrt{2}} \cdot \dfrac{\left(x - 2\right)\left(x + 1\right)}{\left(x - 2\right)}$

$= \dfrac{3}{\sqrt{2}}$.