The value of mathop {lim }limits_{x to 0} fra | vedclass

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(D) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{27^x - 9^x - 3^x + 1}}{{\sqrt 5 - \sqrt {4 + \cos x} }}$.This is a $\frac{0}{0}$ form.Applying $L

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

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(D) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{27^x - 9^x - 3^x + 1}}{{\sqrt 5 - \sqrt {4 + \cos x} }}$.This is a $\frac{0}{0}$ form.Applying $L$-Hospital's rule,we differentiate the numerator and denominator with respect to $x$:Numerator derivative: $\frac{d}{dx}(27^x - 9^x - 3^x + 1) = 27^x \ln 27 - 9^x \ln 9 - 3^x \ln 3 = 3^x \ln 3 (9^x \cdot 3 - 3^x \cdot 2 - 1)$.Denominator derivative: $\frac{d}{dx}(\sqrt 5 - \sqrt {4 + \cos x}) = -\frac{1}{2\sqrt{4 + \cos x}} \cdot (-\sin x) = \frac{\sin x}{2\sqrt{4 + \cos x}}$.Thus,$L = \mathop {\lim }\limits_{x 	o 0} \frac{27^x \ln 27 - 9^x \ln 9 - 3^x \ln 3}{\frac{\sin x}{2\sqrt{4 + \cos x}}} = \mathop {\lim }\limits_{x 	o 0} \frac{2(27^x \ln 27 - 9^x \ln 9 - 3^x \ln 3) \sqrt{4 + \cos x}}{\sin x}$.Using $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$,we multiply and divide by $x$:$L = 2 \sqrt{4+1} \cdot \mathop {\lim }\limits_{x 	o 0} \frac{27^x \ln 27 - 9^x \ln 9 - 3^x \ln 3}{x} = 2\sqrt{5} \cdot (\ln 27 \cdot \ln 27 - \ln 9 \cdot \ln 9 - \ln 3 \cdot \ln 3) = 2\sqrt{5} \cdot (9(\ln 3)^2 - 4(\ln 3)^2 - (\ln 3)^2) = 2\sqrt{5} \cdot 4(\ln 3)^2 = 8\sqrt{5}(\ln 3)^2$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{{{27}^x} - {9^x} - {3^x} + 1}}{{\sqrt 5 - \sqrt {4 + \cos x} }}$ is

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