JEE Main 2020 (Online) 5th September Morning ShiftQuestion 1
If $\alpha$ is the positive root of the equation, $p(x)=x^2-x-2=0$, then $\lim_{x\to\alpha^+}\frac{\sqrt{1-\cos(p(x))}}{x+\alpha-4}$ is equal to :
Limits & Continuity
Solution
$\alpha=2$, $p(x)=(x-2)(x+1)>0$ for $x\to2^+$. $\sqrt{1-\cos p}=\sqrt2|\sin(p/2)|\approx\frac{p}{\sqrt2}$. Denominator $x+2-4=x-2$. Limit $=\frac{(x+1)}{\sqrt2}\to\frac3{\sqrt2}$. Option (C).