Question

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

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

Answer 1

Answer: (C)

$\frac{2}{3}$

Given

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

Applying L'Hopital's rule

$\Rightarrow \underset{x\to 0}{lim}\frac{\frac{6}{(x+3)(-x+3)}}{1}=k$

$\Rightarrow \underset{x\to 0}{lim}\frac{6}{(x+3)(-x+3)}=k$

Putting the value x=0

$\Rightarrow \frac{6}{(0+3)(-0+3)}=k$

$\Rightarrow \frac{6}{9}=k$

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