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(D) We are given the limit $\mathop {\lim }\limits_{x 	o 0} \frac{2}{x}\log (1 + x)$.Using the property of logarithms,$a \log b = \log b^a$,we can re

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$2$

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(D) We are given the limit $\mathop {\lim }\limits_{x 	o 0} \frac{2}{x}\log (1 + x)$.Using the property of logarithms,$a \log b = \log b^a$,we can rewrite the expression as $\mathop {\lim }\limits_{x 	o 0} 2 \log (1 + x)^{\frac{1}{x}}$.Since the logarithm function is continuous,we have $2 \log \left( \mathop {\lim }\limits_{x 	o 0} (1 + x)^{\frac{1}{x}} ight)$.We know that $\mathop {\lim }\limits_{x 	o 0} (1 + x)^{\frac{1}{x}} = e$.Therefore,the expression becomes $2 \log_e e = 2 	imes 1 = 2$.Alternatively,using $L'Hospital's$ rule: $\mathop {\lim }\limits_{x 	o 0} \frac{2 \log(1+x)}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{2 \cdot \frac{1}{1+x}}{1} = \frac{2}{1} = 2$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{2}{x}\log (1 + x)$ is equal to

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