Let f(x) = {(sin x, x ≠ nπ) (2, x = 2), where n ∈ I and g(x) = {x2 + 1, x ≠ 2) (3, x = 2), then lim (x →0) g[f(x)] is
(a) 1
(b) 0
(c) 3
(d) Dose not exist
Correct Option(a) 1
Explanation: