mathop {lim }limits_{x to 0} frac{{x({2^x} - | vedclass

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Question

(B) Given limit: $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x({2^x} - 1)}}{{1 - \cos x}}$Using the standard limits $\mathop {\lim }\limits_{x 	o 0}

Vedclass · Accepted Answer

$\log 4$

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(B) Given limit: $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x({2^x} - 1)}}{{1 - \cos x}}$Using the standard limits $\mathop {\lim }\limits_{x 	o 0} \frac{{{2^x} - 1}}{x} = \log 2$ and $1 - \cos x = 2\sin^2(\frac{x}{2})$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{x({2^x} - 1)}}{{2\sin^2(\frac{x}{2})}}$$L = \mathop {\lim }\limits_{x 	o 0} \left( \frac{{{2^x} - 1}}{x} ight) \cdot \left( \frac{x^2}{2\sin^2(\frac{x}{2})} ight)$$L = (\log 2) \cdot \mathop {\lim }\limits_{x 	o 0} \frac{1}{2 \left( \frac{\sin(x/2)}{x} ight)^2}$Since $\mathop {\lim }\limits_{x 	o 0} \frac{\sin(x/2)}{x/2} = 1$,we have $\mathop {\lim }\limits_{x 	o 0} \frac{\sin(x/2)}{x} = \frac{1}{2}$:$L = (\log 2) \cdot \frac{1}{2 \cdot (1/2)^2} = (\log 2) \cdot \frac{1}{2 \cdot (1/4)} = (\log 2) \cdot 2 = 2\log 2 = \log 4$.

$\mathop {\lim }\limits_{x \to 0} \frac{{x({2^x} - 1)}}{{1 - \cos x}} = $

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