lim _{x rightarrow 0} frac{27^x-9^x-3^x+1}{sq | vedclass

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(C) ધારો કે $L = \lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}$.અંશના અવયવ પાડતા: $27^x-9^x-3^x+1 = (9^x-1)(3^x-1)$.છેદનું સ

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$8 \sqrt{5}(\log 3)^2$

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(C) ધારો કે $L = \lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}$.અંશના અવયવ પાડતા: $27^x-9^x-3^x+1 = (9^x-1)(3^x-1)$.છેદનું સંમેયીકરણ કરતા: $\frac{1}{\sqrt{5}-\sqrt{4+\cos x}} = \frac{\sqrt{5}+\sqrt{4+\cos x}}{1-\cos x}$.તેથી,$L = \lim _{x \rightarrow 0} \frac{(9^x-1)(3^x-1)}{1-\cos x} \cdot (\sqrt{5}+\sqrt{4+\cos x})$.$1-\cos x = 2 \sin^2(x/2)$ નો ઉપયોગ કરતા,$L = \lim _{x \rightarrow 0} \frac{(9^x-1)(3^x-1)}{2 \sin^2(x/2)} \cdot (\sqrt{5}+\sqrt{4+\cos x})$.$x^2$ વડે ગુણતા અને ભાગતા: $L = \lim _{x}$ ${\rightarrow 0} \frac{(\frac{9^x-1}{x})(\frac{3^x-1}{x})}{2 \cdot (\frac{\sin(x/2)}{x/2})^2 \cdot (1/4)} \cdot (\sqrt{5}+\sqrt{4+\cos x})$.$L = \frac{\ln 9 \cdot \ln 3}{1/2} \cdot 2 \sqrt{5} = 8 \sqrt{5} (\ln 3)^2$.

$\lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}=$

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