Question

If $\alpha$ is the positive root of the equation, $p(x)=x^2-x-2=0$, then $\underset{x\to {\alpha }^{+}}{\lim }\frac{\sqrt{1-\cos (p(x))}}{x+\alpha -4}$ is equal to

• A. $\frac{3}{\sqrt{2}}$
• B. $\frac{3}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$x^2-x-2=0$
roots are $2$ & $-1$
$\Rightarrow \underset{x\to 2^{+}}{\lim }\frac{\sqrt{1-\cos \left(x^2-x-2\right)}}{(x-2)}$
$=\underset{x\to 2^{+}}{\lim }\frac{\sqrt{2{\sin }^2\frac{\left(x^2-x-2\right)}{2}}}{(x-2)}$
$=\underset{x\to 2^{+}}{\lim }\frac{\sqrt{2}\sin \left(\frac{(x-2)(x+1)}{2}\right)}{(x-2)}$
$=\frac{3}{\sqrt{2}}$