- A$\frac{1}{3}$
- B$\frac{1}{4}$
- ✓$\frac{1}{6}$
- D$\frac{1}{12}$
$\lim \limits_{x \rightarrow 0}\left(\frac{2 \cdot \sin \left(\frac{x+\sin x}{2}\right) \sin \left(\frac{x-\sin x}{2}\right)}{x^{4}}\right)$
$\lim\limits _{x \rightarrow 0} 2\left(\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\left(\frac{x+\sin x}{2}\right)}\left(\frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}\right)\left(\frac{\left.\frac{x+\sin x}{2}\right)}{x^{4}}\left(\frac{x-\sin x}{2}\right)\right.\right.$
$\lim\limits _{x \rightarrow 0}\left(\frac{x^{2}-\sin ^{2} x}{2 x^{4}}\right):\left(\frac{0}{0}\right)$
Apply $L-Hopital\; Rule$ :
$\lim \limits_{x \rightarrow 0} \frac{2 x-2 \sin x \cos x}{2.4 \cdot x^{3}}$
$\lim\limits _{x \rightarrow 0} \frac{2 x-\sin 2 x}{8 x^{3}} ; \frac{0}{0}:$ Again apply $L-Hopital \;rule$
$\lim \limits_{x \rightarrow 0} \frac{2-2 \cos (2 x)}{8(3) x^{2}}$
$\lim \limits_{x \rightarrow 0} \frac{2(1-\cos (2 x))}{24\left(4 x^{2}\right)} \times 4 \Rightarrow \frac{2}{24} \times \frac{1}{2} \times 4 \Rightarrow \frac{1}{6}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
then the value of $6 \beta$ is $\qquad$