Question

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

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\log \left(3+x\right)-\log \left(3-x\right)}{x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\log \left({\displaystyle \frac{3+x}{3-x}}\right)}{x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\log \left(1+{\displaystyle \frac{3+x}{3-x}}-1\right)}{x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\displaystyle \log \left(1+\frac{3+x-3+x}{3-x}\right)}}{x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\log \left(1+{\displaystyle \frac{2x}{3-x}}\right)}{x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\log \left(1+{\displaystyle \frac{2x}{3-x}}\right)}{\left({\displaystyle \frac{2x}{3-x}}\right)\times {\displaystyle \frac{3-x}{2}}}\right] \\ =\frac{1\times 2}{3-0} \\ =\frac{2}{3} \end{gathered}$$