Question

Discuss the continuity of the function 

f(x) = `("log"(2+"x") - "log"(2-"x"))/("tan""x")` , for x ≠ 0

 = 1  for x = 0 at the point x =0

Answer 1

Here , f(0) = 1    ... (i) [Given]

Also , L = `lim_(x ->0) f("x")`

 

`= lim_(x ->0) f("x") = lim_(x ->0) ("log"(2+"x") - "log"(2-"x"))/("tan""x")`

 

`= lim_(x ->0) ("log"(2 + "x")/(2 - "x"))/("tan""x")       [because "log""m" -"log""n" = "log" ("m"/"n")]`

 

`=lim_(x->0) ("log"((1+"x"/2)/(1-"x"/2)))/("tan""x")`

 

`= lim_(x->0) [["log"((1+"x"/2)/(1-"x"/2)))/"x"] xx lim_(x->0) "x"/("tan""x")`

`= [lim_(x->0) ("log" (1 + "x"/2))/"x" - lim_(x->0) ("log" (1 - "x"/2))/"x"] xx lim_(x->0) "x"/"tan x"`

`= [1/2 - ((-1)/2)]xx 1`

= 1

From (i) and (ii), we have `lim_(x->0) "f"("x") = "f"(0)`

∴ The given function is continuous at x = 0