Discuss the continunity of the funcation
`f(x) = (log (2 + x) - (log (2 - x)))/(tan x ) "for x " ne 0 `
= 1 for x = 0
at the point x= 0

Explanation

f (x) is continuous at x = 0 if ` underset(x to 0 ) ("lim") f (x) = f(0) `. But ,
` R.H.S = f (0) = 1 " "` …(Given)
L.H.S `= underset(x to 0 ) (lim) f (x) `
`= underset(x to 0) lim (log (2+ x ) - log (2 - x))/(tan x )`
` = underset(x to 0)lim (log ((2 + x)/(2 -x)))/(tan x)`
` = underset(x to 0)lim (1)/(x) (log ((2 + x)/(2 -x)))/((tan x)/(x))`
... (Dividing both Nr and Dr by x )
` = underset(x to 0)lim (log ((2 + x)/(2 -x))^(1//x))/((tan x)/(x))`
` = underset(x to 0)lim (log [(1 + (x)/(2))/(1 - (x)/(2))])/((tan x//x))`
` underset(x to 0)lim (log[({(1 + (x)/(2))^(1//2)})/({(1 - (x)/(2))^(-1//2)})])/(underset(x to 0)lim ((tanx)/(x)))`
` therefore L.H.S = (log ((e^(1//2))/(e^(-1//2))))/(1) = log e^(1//2 + 1//2)`
= log e = 1
` therefore` L.H.S . = R.H.S
hence , f(x) is continuous at x = 0 .