Question

If $\underset{x\to 0}{lim}\frac{\log (3+x)-\log (3-x)}{x}=k$ then k=?

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

Answer 1

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

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

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

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

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

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

$k=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

Hence, option 'B' is correct.