Question

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

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

Answer 1

Answer: (A)

$\frac{2}{3}$

Given, $\underset{x\to 0}{lim}\frac{\ln (3+x)-\ln (3-x)}{x}=k$

Applying L'Hospital's rule
${lim}_{x\to 0}\frac{\log (3+x)-\log (3-x)}{x}$

$={lim}_{x\to 0}\frac{\frac{1}{3+x}+\frac{1}{3-x}}{1}=\frac{2}{3}$

$\therefore k=\frac{2}{3}$