lim _{x rightarrow 0} frac{sqrt{2}-sqrt{1+cos | vedclass

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(A) ધારો કે $L = \lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2x}-4}$.નિત્યસમ $1+\cos x = 2\cos^2(x/2)$ અને $\cos 2x = 2\cos^

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) ધારો કે $L = \lim _{x ightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2x}-4}$.નિત્યસમ $1+\cos x = 2\cos^2(x/2)$ અને $\cos 2x = 2\cos^2 x - 1$ નો ઉપયોગ કરતા:$L = \lim _{x ightarrow 0} \frac{\sqrt{2}-\sqrt{2}\cos(x/2)}{\sqrt{14+2\cos^2 x}-4}$.અંશ અને છેદનું સંમેયીકરણ કરતા:$L = \lim _{x ightarrow 0} \frac{\sqrt{2}(1-\cos(x/2))}{\sqrt{14+2\cos^2 x}-4} 	imes \frac{1+\cos(x/2)}{1+\cos(x/2)} 	imes \frac{\sqrt{14+2\cos^2 x}+4}{\sqrt{14+2\cos^2 x}+4}$.$L = \lim _{x ightarrow 0} \frac{\sqrt{2}\sin^2(x/2)}{1+\cos(x/2)} 	imes \frac{\sqrt{14+2\cos^2 x}+4}{14+2\cos^2 x - 16}$.કારણ કે $14+2\cos^2 x - 16 = 2\cos^2 x - 2 = -2\sin^2 x = -8\sin^2(x/2)\cos^2(x/2)$:$L = \lim _{x ightarrow 0} \frac{\sqrt{2}\sin^2(x/2)}{1+\cos(x/2)} 	imes \frac{\sqrt{14+2\cos^2 x}+4}{-8\sin^2(x/2)\cos^2(x/2)}$.$L = \frac{\sqrt{2}}{1+1} 	imes \frac{\sqrt{14+2}+4}{-8(1)} = \frac{\sqrt{2}}{2} 	imes \frac{8}{-8} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$.

$\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2x}-4} = $

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