Mathematics - Limits Question with Solution | TestHub

Given, $p\left(x\right)=0$

$\Rightarrow x^2-x-2=0$

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

$\Rightarrow x=2,-1 \therefore \alpha =2$

Now,$\underset{x\to 2^{+}}{\lim }\frac{\sqrt{1-\cos \left(x^2-x-2\right)}}{x-2}$

$=\underset{x\to 2^{+}}{\lim }\frac{\sqrt{2{\sin }^2\left(\frac{x^2-x-2}{2}\right)}}{x-2}$

$=\underset{x\to 2^{+}}{\lim }\sqrt{2}·\frac{\left|\sin \left(\frac{x^2-x-2}{2}\right)\right|}{x-2}$

$=\underset{x\to 2^{+}}{\lim }\sqrt{2}·\frac{\sin \left(\frac{x^2-x-2}{2}\right)}{x-2}$ (because $x>2$)

$=\underset{x\to 2^{+}}{\lim } \frac{\sqrt{2}\sin \left(\frac{x^2-x-2}{2}\right)}{\left(\frac{x^2-x-2}{2}\right)·2}·\left(\frac{x^2-x-2}{x-2}\right)$

$=\underset{x\to 2^{+}}{\lim }\frac{1}{\sqrt{2}}·\frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)}$

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