Question

Evaluate:
$\underset{x\to 0}{lim}\frac{\log (2+x)-\log (2-x)}{x}$

Answer 1

$\underset{x\to 0}{lim}\frac{log(2+x)-log(2-x)}{x}$

$=\underset{x\to 0}{lim}\frac{log\left(\frac{2+x}{2-x}\right)}{x}$ Adding and substracting $1$,

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

$=\underset{x\to 0}{lim}\frac{log(1+\frac{2x}{2-x})}{\left(\frac{2x}{2-x}\right)\times \left(\frac{2-x}{2}\right)}$

$=\underset{x\to 0}{lim}\frac{log[1+\frac{2x}{2-x}]}{\frac{2x}{2-x}}\times \underset{x\to 0}{lim}\frac{2}{2-x}$ $(\because \underset{x\to 0}{lim}\frac{log(1+x)}{x}=1)$

$=1\times \frac{2}{2-0}=1$.