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

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(A) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(2^{x}-1)}{1-\cos x}$ Since this is a $\frac{0}{0}$ form,we apply $L$'Hopital's rule: $L = \lim _

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$2 \log 2$

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(A) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(2^{x}-1)}{1-\cos x}$ Since this is a $\frac{0}{0}$ form,we apply $L$'Hopital's rule: $L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(x 2^{x}-x)}{\frac{d}{dx}(1-\cos x)}$ $L = \lim _{x \rightarrow 0} \frac{2^{x} + x 2^{x} \ln 2 - 1}{\sin x}$ This is still a $\frac{0}{0}$ form. Applying $L$'Hopital's rule again: $L = \lim _{x \rightarrow 0} \frac{2^{x} \ln 2 + 2^{x} \ln 2 + x 2^{x} (\ln 2)^{2}}{\cos x}$ Substituting $x = 0$: $L = \frac{2^{0} \ln 2 + 2^{0} \ln 2 + 0}{\cos 0} = \frac{\ln 2 + \ln 2}{1} = 2 \ln 2$ Thus,the correct option is $A$.

$\lim _{x \rightarrow 0} \frac{x 2^{x}-x}{1-\cos x}$ is equal to

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