Question

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

Answer 1

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

$={lim}_{x\to 0}\frac{log\left(\frac{3+x}{3-x}\right)}{x}$

$={lim}_{x\to 0}\frac{log\left[\frac{(3-x)+2x}{3-x}\right]}{x}$

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

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

$=\frac{2}{3}$