Question

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

Answer 1

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

Here $f\left(x\right)=\log (3-x)-\log (3+x)$

$x\to$ $f\left(0\right)=\log 3-\log 3=0$

and $g\left(x\right)=x$

$x\to 0,g\left(0\right)=0$

$\frac{f\left(0\right)}{g\left(0\right)}=\frac{0}{0}$

Byt using L Hospitals rule

${lim}_{x\to 0}\frac{f\left(x\right)}{g\left(x\right)}={lim}_{x\to 0}\frac{\frac{1}{3+x}+\frac{1}{3-x}}{1}$

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