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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
Solution
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
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