lim_{x rightarrow frac{pi}{2}} (tan^{2} x (sq | vedclass

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Question

(A) ધારો કે $L = \lim_{x ightarrow \frac{\pi}{2}} 	an^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2})$.લક્ષની અંદરની

Vedclass · Accepted Answer

$\frac{1}{12}$

Vedclass · Answer

(A) ધારો કે $L = \lim_{x ightarrow \frac{\pi}{2}} 	an^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2})$.લક્ષની અંદરની પદાવલિનું સંમેયીકરણ કરતા:$L = \lim_{x ightarrow \frac{\pi}{2}} 	an^{2} x \cdot \frac{(2 \sin^{2} x + 3 \sin x + 4) - (\sin^{2} x + 6 \sin x + 2)}{\sqrt{2 \sin^{2} x + 3 \sin x + 4} + \sqrt{\sin^{2} x + 6 \sin x + 2}}$$L = \lim_{x ightarrow \frac{\pi}{2}} 	an^{2} x \cdot \frac{\sin^{2} x - 3 \sin x + 2}{\sqrt{2 \sin^{2} x + 3 \sin x + 4} + \sqrt{\sin^{2} x + 6 \sin x + 2}}$જેમ $x ightarrow \frac{\pi}{2}$,તેમ $\sin x ightarrow 1$. છેદ $\sqrt{2+3+4} + \sqrt{1+6+2} = \sqrt{9} + \sqrt{9} = 6$ ને અનુલક્ષે છે.$L = \frac{1}{6} \lim_{x ightarrow \frac{\pi}{2}} 	an^{2} x (\sin x - 1)(\sin x - 2)$કારણ કે $\sin x - 2 ightarrow -1$ જેમ $x ightarrow \frac{\pi}{2}$,તેથી:$L = \frac{1}{6} \lim_{x ightarrow \frac{\pi}{2}} \frac{\sin^{2} x}{\cos^{2} x} (\sin x - 1)(-1) = \frac{1}{6} \lim_{x ightarrow \frac{\pi}{2}} \frac{\sin^{2} x (1 - \sin x)}{1 - \sin^{2} x}$$L = \frac{1}{6} \lim_{x ightarrow \frac{\pi}{2}} \frac{\sin^{2} x (1 - \sin x)}{(1 - \sin x)(1 + \sin x)} = \frac{1}{6} \cdot \frac{1^2}{1+1} = \frac{1}{12}$.

$\lim_{x \rightarrow \frac{\pi}{2}} (\tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2}))$ ની કિંમત શોધો.

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