Question

The value of $\underset{x\to 0}{lim}\frac{2{\int }_1^{\cos x}{\sin }^{-1}\left(\sqrt{1-t^2}\right)dt}{2x-\sin 2x}$ is

• A. $0$
• B. does not exist
• C. $-1$
• D. $\frac{-1}{2}$

Answer 1

The correct option is B does not exist

$\underset{x\to 0}{lim}\frac{2{\int }_1^{\cos x}{\sin }^{-1}\left(\sqrt{1-t^2}\right)dt}{2x-\sin 2x}$
On applying $\underset{x\to 0}{lim},$ we get $\frac{0}{0}$ form.
Therefore, use L'Hospitals' rule,
$\underset{x\to 0}{lim}\frac{2\left({\sin }^{-1}\sqrt{1-{\cos }^{2}x}\right)(-\sin x)}{2-2\cos 2x}$
$=\underset{x\to 0}{lim}\frac{-\left({\sin }^{-1}\right|\sin x\left|\right)\sin x}{2{\sin }^{2}x}$

$=\underset{x\to 0}{lim}\frac{-{\sin }^{-1}|\sin x|}{2\sin x}$
L.H.L $=\underset{x\to 0^{-}}{lim}\frac{-(-x)}{2\sin x}=\frac{1}{2}$
R. H. L $=\underset{x\to 0^{+}}{lim}\frac{-\left(x\right)}{2\sin x}=\frac{-1}{2}.$
L. H. L $\ne$ R.H.L
$\therefore$ limit does not exist.