Question

Discuss the continuity of $f\left(x\right)=\frac{log(2+x)-log(2-x)}{tanx},forx\ne 0=1,forx=0.$
at $x=0$

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{\log (2+x)-\log (2-x)}{\tan x},x\ne 0\\ 1,x=0\end{array}$

Checking continuity at $x=0$

$\Rightarrow {lim}_{x\to 0^{+}}f\left(x\right)={lim}_{h\to 0}\frac{\log (2+0+h)-\log (2-0-h)}{\tan (0+h)}={lim}_{h\to 0}\frac{\frac{1}{2+h}-\frac{(-1)}{2-h}}{{\sec }^{2}h}=1\to \left(1\right)$ (By L' Hospital Rule)

$\Rightarrow {lim}_{x\to 0^{-}}f\left(x\right)={lim}_{h\to 0}\frac{\log (2+0-h)-\log (2-0+h)}{\tan (0-h)}={lim}_{h\to 0}\frac{\frac{-1}{2-h}-\frac{1}{2+h}}{-{\sec }^{2}h}=1\to \left(2\right)$ (By L' Hospital Rule)

$\Rightarrow f\left(0\right)=1$ $\to \left(3\right)$

From $1,2\&3$ we get

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

$\Rightarrow f\left(x\right)$ is continuous at $x=0$