If f x = e 2 x + 2 x + 1 - 1 2 x + 2 x + 1 : x ≠ 0 1 : x = 0 , then (where, . represents the greatest integer function)

Step-by-step solution

lim x → 0 + ⁡ f x ⁡ = lim h → 0 + ⁡ f 0 + h = lim h → 0 + ⁡ e 2 h + 2 h + 1 - 1 2 h + 2 h + 1 = lim h → 0 + ⁡ e 2 h + 1 - 1 2 h + 1 = e - 1 And, lim x → 0 - ⁡ f x = lim h → 0 + ⁡ f 0 - h = lim h → 0 + ⁡ e - 2 h - 2 h + 1 - 1 - 2 h - 2 h + 1 = lim h → 0 + ⁡ e - 2 h - 1 - 2 h = 1 Since, lim x → 0 + ⁡ f x ⁡ ≠ lim x → 0 - ⁡ f x , therefore, f x is not continuous at x = 0