Question

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

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

Answer 1

The correct option is D $\frac{3}{\sqrt{2}}$

$f\left(x\right)=x^2-x-2=0\Rightarrow \alpha =2$

$L=\underset{x\to 2^{+}}{lim}\frac{\sqrt{1-\cos (x-2)(x+1)}}{x+2-4}$
$=\underset{x\to 2^{+}}{lim}\frac{\sqrt{1-\cos (x-2)(x+1)}}{(x-2)}$
$=\underset{h\to 0}{lim}\frac{\sqrt{1-\cos (h\times (h+3))}}{h}$
$=\underset{h\to 0}{lim}\sqrt{\frac{1-\cos \left(h(h+3)\right)}{h^2{(h+3)}^2}\times {(h+3)}^2}$
$=\sqrt{\frac{1}{2}\times 9}=\frac{3}{\sqrt{2}}$