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If $\lim _{x \rightarrow 0} \frac{\log (3+x)-\log (3-x)}{x}=k$, the value of $k$ is
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The correct answer is:
$\frac{2}{3}$
$\frac{2}{3}$
$\operatorname{Lim}_{x \rightarrow 0} \frac{\log (3+x)-\log (3-x)}{x}=K$ (by L'Hospital rule)
$\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{1}{3+x}-\frac{-1}{3-x}}{1}=K \quad \therefore \frac{2}{3}=\mathrm{K}$
$\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{1}{3+x}-\frac{-1}{3-x}}{1}=K \quad \therefore \frac{2}{3}=\mathrm{K}$
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