limx→0x(log(1+x2)-log(x2))

limx→0 x log 1+x2-logx2 =limx→0 x log 1+x2x2 =limx→0 2x2 log1+1x2 =limx→02log1+1x2x2 =2 logee =2

limx→0x(log(1+x2)-log(x2))

limx→0 x log 1+x2-logx2 =limx→0 x log 1+x2x2 =limx→0 2x2 log1+1x2 =limx→02log1+1x2x2 =2 logee =2

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$\lim_{x \to 0} x (\log (1 + \frac{x}{2}) - \log (\frac{x}{2}))$

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answer is C.

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Detailed Solution

$\lim_{x \to 0} x [\log (1 + \frac{x}{2}) - \log \frac{x}{2}] = \lim_{x \to 0} x \log (\frac{1 + \frac{x}{2}}{\frac{x}{2}}) = \lim_{x \to 0} 2 \frac{x}{2} \log (1 + \frac{1}{\frac{x}{2}}) = \lim_{x \to 0} 2 \log {(1 + \frac{1}{\frac{x}{2}})}^{\frac{x}{2}} = 2 {\log_{e}}^{e} = 2$