Question

Solve by using L- Hospital Rule : $\underset{x⟶0}{lim}\frac{\log (3+x)-\log (3-x)}{x}$

Answer 1

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

[Applying L-hospital rule]

$={lim}_{x\to 0}\frac{\frac{1}{3+x}.1-\frac{1}{3-x}(-1)}{1}={lim}_{x\to 0}\frac{\frac{1}{3+x}+\frac{1}{3-x}}{1}={lim}_{x\to 0}\frac{3-x+3+x}{(3+x)(3-x))}={lim}_{x\to 0}\frac{6}{9-0^2}=\frac{6}{9}=\frac{2}{3}$