Question

If $\underset{x\to 0}{lim}\frac{\log (3+x)-\log (3-x)}{x}=k$, the value of $k$ is

• A. $-\frac{2}{3}$
• B. $0$
• C. $-\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Substituting gives indeterminate form.
So, applying L'Hospital's Rule,
$={lim}_{x\to 0}(\frac{1}{3+x}+\frac{1}{3-x})$
$=\frac{1}{3}+\frac{1}{3}$
$=\frac{2}{3}$
Hence, option 'D' is correct.