The value of the limit lim_{x rightarrow frac | vedclass

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(C) Let the given limit be $L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{

Vedclass · Accepted Answer

$8$

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(C) Let the given limit be $L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$.Using the identity $\sin 3x + \sin x = 2 \sin 2x \cos x$,the numerator becomes $8 \sqrt{2} \sin 2x \cos x = 16 \sqrt{2} \sin x \cos^2 x$.For the denominator,we use $\cos \frac{5x}{2} - \cos \frac{3x}{2} = -2 \sin 2x \sin \frac{x}{2}$ and $\sqrt{2} + \sqrt{2} \cos 2x = \sqrt{2}(1 + \cos 2x) = 2 \sqrt{2} \cos^2 x$.The denominator simplifies to $2 \sin 2x \sin \frac{3x}{2} - 2 \sin 2x \sin \frac{x}{2} - 2 \sqrt{2} \cos^2 x = 2 \sin 2x (\sin \frac{3x}{2} - \sin \frac{x}{2}) - 2 \sqrt{2} \cos^2 x$.Using $\sin \frac{3x}{2} - \sin \frac{x}{2} = 2 \cos x \sin \frac{x}{2}$,the denominator becomes $4 \sin x \cos x (2 \cos x \sin \frac{x}{2}) - 2 \sqrt{2} \cos^2 x = 2 \cos^2 x (4 \sin x \sin \frac{x}{2} - \sqrt{2})$.Thus,$L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{16 \sqrt{2} \sin x \cos^2 x}{2 \cos^2 x (4 \sin x \sin \frac{x}{2} - \sqrt{2})} = \lim_{x \rightarrow \frac{\pi}{2}} \frac{8 \sqrt{2} \sin x}{4 \sin x \sin \frac{x}{2} - \sqrt{2}}$.Substituting $x = \frac{\pi}{2}$,we get $L = \frac{8 \sqrt{2} (1)}{4 (1) \sin \frac{\pi}{4} - \sqrt{2}} = \frac{8 \sqrt{2}}{4 \cdot \frac{1}{\sqrt{2}} - \sqrt{2}} = \frac{8 \sqrt{2}}{2 \sqrt{2} - \sqrt{2}} = \frac{8 \sqrt{2}}{\sqrt{2}} = 8$.

The value of the limit $\lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$ is

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