If limx→0log3+x-log3-xx=k, the value of k is

ltx→0log(3+x)-log(3-x)x 00 form =ltx→013+x-13-x(-1) (L'Hospital Rule) =13+13=23

If limx→0log3+x-log3-xx=k, the value of k is

ltx→0log(3+x)-log(3-x)x 00 form =ltx→013+x-13-x(-1) (L'Hospital Rule) =13+13=23

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If $\lim_{x \to 0} \frac{\log (3 + x) - \log (3 - x)}{x} = k,$ the value of k is

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$\frac{2}{3}$

$- \frac{2}{3}$

$- \frac{1}{3}$

answer is D.

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$\underset{x \to 0}{lt} \frac{\log (3 + x) - \log (3 - x)}{x} (\frac{0}{0} form) = \underset{x \to 0}{lt} \frac{1}{3 + x} - \frac{1}{3 - x} (- 1) (L^{'} Hospital Rule) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

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If limx→0log3+x-log3-xx=k, the value of k is

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If $\lim_{x \to 0} \frac{\log (3 + x) - \log (3 - x)}{x} = k,$ the value of k is