લક્ષ lim_{x rightarrow frac{pi}{2}} frac{4 sq | vedclass

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Question

(C) ધારો કે આપેલ લક્ષ $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

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) ધારો કે આપેલ લક્ષ $L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$ છે.નિત્યસમ $\sin 3x + \sin x = 2 \sin 2x \cos x$ નો ઉપયોગ કરતા,અંશ $8 \sqrt{2} \sin 2x \cos x = 16 \sqrt{2} \sin x \cos^2 x$ બને છે.છેદ માટે,આપણે $\cos \frac{5x}{2} - \cos \frac{3x}{2} = -2 \sin 2x \sin \frac{x}{2}$ અને $\sqrt{2} + \sqrt{2} \cos 2x = 2 \sqrt{2} \cos^2 x$ નો ઉપયોગ કરીએ છીએ.છેદનું સાદું રૂપ $2 \sin 2x (\sin \frac{3x}{2} - \sin \frac{x}{2}) - 2 \sqrt{2} \cos^2 x$ થાય છે.$\sin \frac{3x}{2} - \sin \frac{x}{2} = 2 \cos x \sin \frac{x}{2}$ નો ઉપયોગ કરતા,છેદ $2 \cos^2 x (4 \sin x \sin \frac{x}{2} - \sqrt{2})$ બને છે.આમ,$L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{8 \sqrt{2} \sin x}{4 \sin x \sin \frac{x}{2} - \sqrt{2}} = 8$.

લક્ષ $\lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$ ની કિંમત શોધો.

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