The value of the limit mathop {lim }limits_{x | vedclass

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(C) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$Since the form is $\frac{0}{0}$,we apply $L'	ext{Ho

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$2$

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(C) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$Since the form is $\frac{0}{0}$,we apply $L'	ext{Hospital's rule}$:$= \mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}({e^x} - {e^{ - x}} - 2x)}{\frac{d}{dx}(x - \sin x)} = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + {e^{ - x}} - 2}}{{1 - \cos x}}$Applying $L'	ext{Hospital's rule}$ again as it is still $\frac{0}{0}$:$= \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$Applying $L'	ext{Hospital's rule}$ for the third time:$= \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + {e^{ - x}}}}{{\cos x}} = \frac{{{e^0} + {e^0}}}{{\cos 0}} = \frac{{1 + 1}}{1} = 2$

The value of the limit $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$ is

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