Iflimx→0⁡log⁡(3+x)−log⁡(3−x)x=k then k=

k=Ltx→0log3+x−log3−xx=Ltx→013+x+13−x=13+13=23 by LH rule

Iflimx→0⁡log⁡(3+x)−log⁡(3−x)x=k then k=

k=Ltx→0log3+x−log3−xx=Ltx→013+x+13−x=13+13=23 by LH rule

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$If \lim_{x \to 0} \frac{\log (3 + x) - \log (3 - x)}{x} = k then k =$

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$\frac{2}{3}$

$- \frac{2}{3}$

$- \frac{1}{3}$

answer is C.

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$k = \underset{x \to 0}{L t} \frac{\log (3 + x) - \log (3 - x)}{x} = \underset{x \to 0}{L t} \frac{1}{3 + x} + \frac{1}{3 - x} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ $(b y L H r u l e)$

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Iflimx→0⁡log⁡(3+x)−log⁡(3−x)x=k then k=

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$If \lim_{x \to 0} \frac{\log (3 + x) - \log (3 - x)}{x} = k then k =$