If f(x) = frac{x(a^x - 1)}{1 - cos x} and g(x | vedclass

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(D) First,evaluate $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{x(a^x - 1)}{1 - \cos x}$.Using the standard limits $\lim_{x 	o 0} \frac{a^x - 1}{x} =

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$\log a$

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(D) First,evaluate $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{x(a^x - 1)}{1 - \cos x}$.Using the standard limits $\lim_{x 	o 0} \frac{a^x - 1}{x} = \log a$ and $\lim_{x 	o 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$,we rewrite $f(x)$ as $\frac{(a^x - 1)/x}{(1 - \cos x)/x^2} = \frac{\log a}{1/2} = 2 \log a$.Next,evaluate $\lim_{x 	o 0} g(x) = \lim_{x 	o 0} \frac{x(1 - a^x)}{a^x(\sqrt{1 - x^2} - \sqrt{1 + x^2})}$.Rationalize the denominator: $\sqrt{1 - x^2} - \sqrt{1 + x^2} = \frac{(1 - x^2) - (1 + x^2)}{\sqrt{1 - x^2} + \sqrt{1 + x^2}} = \frac{-2x^2}{\sqrt{1 - x^2} + \sqrt{1 + x^2}}$.So,$g(x) = \frac{x(1 - a^x)}{a^x \cdot \frac{-2x^2}{\sqrt{1 - x^2} + \sqrt{1 + x^2}}} = \frac{-(a^x - 1)}{x} \cdot \frac{\sqrt{1 - x^2} + \sqrt{1 + x^2}}{-2 a^x} = \frac{a^x - 1}{x} \cdot \frac{\sqrt{1 - x^2} + \sqrt{1 + x^2}}{2 a^x}$.Taking the limit as $x 	o 0$: $\lim_{x 	o 0} g(x) = (\log a) \cdot \frac{1 + 1}{2(1)} = \log a$.Finally,$\lim_{x 	o 0} (f(x) - g(x)) = 2 \log a - \log a = \log a$.

If $f(x) = \frac{x(a^x - 1)}{1 - \cos x}$ and $g(x) = \frac{x(1 - a^x)}{a^x(\sqrt{1 - x^2} - \sqrt{1 + x^2})}$,then $\lim_{x \to 0} (f(x) - g(x)) = $

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