$\lim\limits_{x \rightarrow 0} \frac{2 \int\limits_0^{\cos x} \cos ^{-1} t d t}{2 x-\sin 2 x}$ is equal to
Answer & explanation
Correct answer: option 3
We have,
$\lim\limits_{x \rightarrow 0} \frac{2 \int\limits_0^{\cos x} \cos ^{-1} t d t}{2 x-\sin 2 x}$
$=\lim\limits_{x \rightarrow 0} \frac{-2 \sin x \cos ^{-1}(\cos x)}{2-2 \cos 2 x}$ [Applying L'Hospital's rule]
$= -\lim\limits_{x \rightarrow 0} \frac{x \sin x}{2 \sin ^2 x}=-\frac{1}{2} \lim\limits_{x \rightarrow 0} \frac{x}{\sin x}=-\frac{1}{2}$