A) 0
B) \[-\frac{1}{3}\]
C) \[\frac{2}{3}\]
D) \[-\frac{2}{3}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (3+x)-\log (3-x)}{x}=k\] By L-Hospital?s rule, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{3+x}+\frac{1}{3-x}}{1}=k\] Þ \[\frac{2}{3}=k\].
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