Question

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

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

Answer 1

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

$\underset{x\to 0}{lim}\frac{\log (3+x)-\log (3-x)}{x}$

$=\underset{x\to 0}{lim}\frac{\log 3+\log (1+\frac{x}{3})-\log 3-\log (1-\frac{x}{3})}{x}$

$=\underset{x\to 0}{lim}\frac{\log (1+\frac{x}{3})-\log (1-\frac{x}{3})}{x}=\underset{x\to 0}{lim}\frac{(\frac{x}{3}-\frac{x^2}{2.3^2})-(-\frac{x}{3}-\frac{x^2}{2.3^2})}{x}$

$=\underset{x\to 0}{lim}\frac{\frac{2x}{3}+\frac{2x^3}{3^3.3}}{x}=\frac{2}{3}$