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Question: If\(\underset{\mathbf{x}\mathbf{\rightarrow}\mathbf{0}}{\mathbf{\lim}}\frac{\mathbf{\log}\mathbf{(}\...

Iflim⁡x→0log⁡(3+x)−log⁡(3−x)x=k,\underset{\mathbf{x}\mathbf{\rightarrow}\mathbf{0}}{\mathbf{\lim}}\frac{\mathbf{\log}\mathbf{(}\mathbf{3 + x)}\mathbf{-}\mathbf{\log}\mathbf{(}\mathbf{3}\mathbf{-}\mathbf{x)}}{\mathbf{x}}\mathbf{= k,} then the value of k is

A

0

B

−13- \frac{1}{3}

C

23\frac{2}{3}

D

−23- \frac{2}{3}

Answer

23\frac{2}{3}

Explanation

Solution

lim⁡x→0log⁡(3+x)−log⁡(3−x)x=lim⁡x→0log⁡(3+x3−x)x=lim⁡x→0log⁡(1+(x/3)1−(x/3))x=lim⁡x→0log⁡(1+(x/3))x−lim⁡x→0log⁡(1−(x/3))x=13−(−13)=23\lim_{x \rightarrow 0}\frac{\log(3 + x) - \log(3 - x)}{x} = \lim_{x \rightarrow 0}\frac{\log\left( \frac{3 + x}{3 - x} \right)}{x} = \lim_{x \rightarrow 0}\frac{\log\left( \frac{1 + (x/3)}{1 - (x/3)} \right)}{x} = \lim_{x \rightarrow 0}\frac{\log\left( 1 + (x/3) \right)}{x} - \lim_{x \rightarrow 0}\frac{\log\left( 1 - (x/3) \right)}{x} = \frac{1}{3} - \left( - \frac{1}{3} \right) = \frac{2}{3}.