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

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(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{x({e^x} - 1)}}{{1 - \cos x}}$.Using the standard limits $\mathop {\lim }\l

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$2$

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(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{x({e^x} - 1)}}{{1 - \cos x}}$.Using the standard limits $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - 1}}{x} = 1$ and the identity $1 - \cos x = 2 \sin^2(x/2)$:$\mathop {\lim }\limits_{x 	o 0} \frac{{x({e^x} - 1)}}{{2 \sin^2(x/2)}}$.Multiply and divide by $x$ and $(x/2)^2$ to use the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$:$= \mathop {\lim }\limits_{x 	o 0} \left( \frac{{{e^x} - 1}}{x} ight) \cdot \frac{x^2}{2 \sin^2(x/2)}$.$= 1 \cdot \mathop {\lim }\limits_{x 	o 0} \frac{x^2}{2 (x/2)^2 \cdot (\frac{\sin(x/2)}{x/2})^2}$.$= \mathop {\lim }\limits_{x 	o 0} \frac{x^2}{2 (x^2/4) \cdot 1^2} = \frac{1}{2/4} = 2$.

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

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