lim _{x rightarrow 0} operatorname{cosec} xle | vedclass

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Question

(D) ધારો કે $L = \lim _{x \rightarrow 0} \frac{\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}}{\sin x}$.અંશનું સંમેયીકરણ કરતા:$L = \lim _{x}$ $

Vedclass · Accepted Answer

$-\frac{1}{2 \sqrt{5}}$

Vedclass · Answer

(D) ધારો કે $L = \lim _{x \rightarrow 0} \frac{\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}}{\sin x}$.અંશનું સંમેયીકરણ કરતા:$L = \lim _{x}$ ${\rightarrow 0} \frac{(2 \cos ^2 x+3 \cos x) - (\cos ^2 x+\sin x+4)}{\sin x (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$$L = \lim _{x}$ ${\rightarrow 0} \frac{\cos ^2 x+3 \cos x - \sin x - 4}{\sin x (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$$\cos ^2 x + 3 \cos x - 4 = (\cos x - 1)(\cos x + 4)$ હોવાથી:$L = \lim _{x}$ ${\rightarrow 0} \frac{(\cos x - 1)(\cos x + 4) - \sin x}{\sin x (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$$\cos x - 1 = -2 \sin ^2 \frac{x}{2}$ અને $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ નો ઉપયોગ કરતા:$L = \lim _{x}$ ${\rightarrow 0} \frac{-2 \sin ^2 \frac{x}{2}(\cos x + 4) - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2} (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$અંશ અને છેદને $2 \sin \frac{x}{2}$ વડે ભાગતા:$L = \lim _{x}$ ${\rightarrow 0} \frac{-\sin \frac{x}{2}(\cos x + 4) - \cos \frac{x}{2}}{\cos \frac{x}{2} (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$જ્યારે $x \rightarrow 0$,ત્યારે $\sin \frac{x}{2} \rightarrow 0$,$\cos \frac{x}{2} \rightarrow 1$,$\cos x \rightarrow 1$:$L = \frac{-(0)(5) - 1}{1 (\sqrt{5} + \sqrt{5})} = -\frac{1}{2 \sqrt{5}}$.

$\lim _{x \rightarrow 0} \operatorname{cosec} x\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right)$ ની કિંમત શોધો.

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