Let f(x) = frac{x cdot 2^x - x}{1 - cos x} an | vedclass

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(D) For $f(x) = \frac{x(2^x - 1)}{1 - \cos x} = \frac{x(2^x - 1)}{2 \sin^2(x/2)}$.As $x 	o 0$,$f(x) = \frac{x(2^x - 1)}{2 (x/2)^2} = \frac{x(2^x - 1)

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$(B)$ and $(C)$ are both correct

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(D) For $f(x) = \frac{x(2^x - 1)}{1 - \cos x} = \frac{x(2^x - 1)}{2 \sin^2(x/2)}$.As $x 	o 0$,$f(x) = \frac{x(2^x - 1)}{2 (x/2)^2} = \frac{x(2^x - 1)}{x^2/2} = 2 \frac{2^x - 1}{x}$.Using the standard limit $\lim_{x 	o 0} \frac{a^x - 1}{x} = \ln a$,we get $\lim_{x 	o 0} f(x) = 2 \ln 2 = \ln 4$.Thus,option $(C)$ is correct.For $g(x) = 2^x \sin \left( \frac{\ln 2}{2^x} ight)$,as $x 	o \infty$,$2^x 	o \infty$,so $\frac{\ln 2}{2^x} 	o 0$.Let $u = \frac{\ln 2}{2^x}$,then $g(x) = \frac{\ln 2}{u} \sin(u) = \ln 2 \cdot \frac{\sin u}{u}$.As $x 	o \infty$,$u 	o 0$,so $\lim_{x 	o \infty} g(x) = \ln 2 \cdot 1 = \ln 2$.Thus,option $(B)$ is correct.Therefore,$(D)$ is the correct option.

Let $f(x) = \frac{x \cdot 2^x - x}{1 - \cos x}$ and $g(x) = 2^x \sin \left( \frac{\ln 2}{2^x} \right)$,then:

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