A) 0
B) -1/3
C) 2/3
D) -2/3
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\log \,(3+x)-\log \,(3-x)}{x}=k\] Applying L' Hospital rule, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{1}{3+x}+\frac{1}{3-x} \right)}{1}=k\] \[\Rightarrow \] \[\frac{1}{3}+\frac{1}{3}=k\] \[\Rightarrow \] \[k=\frac{2}{3}\] NOTE L' Hospital rule can be applied only, if a function is of the form of \[\left( \frac{0}{0} \right)\] or \[\left( \frac{\infty }{\infty } \right)\] form.
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