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Q. $\underset{x \rightarrow 0}{\text{lt}}\left[\frac{\log (3+x)-\log (3-x)}{x}\right]=....$

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Solution:

By L.H. Rule, given $=\underset{x \rightarrow 0}{\text{lt}}\left[\frac{\frac{1}{3+x}-\frac{1}{3-x}}{1}\right]=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$