Let f : R → R be a function such that f x ≤ x 2 , for all x ∈ R . Then, at x = 0 , f is

Step-by-step solution

We know that if y ≤ k 2 ,     ⇒ - k ≤ y ≤ k ,   k ∈ R . Given f x ≤ x 2 ,   ∀   x ∈ R ⇒ - x 2 ≤ f x ≤ x 2   ∀   x ∈ R Now, lim x → 0 - x 2 = lim x → 0 x 2 = 0 , Hence, by using sandwich theorem, lim x → 0 f x = 0 Further f 0 ≤ 0   ⇒ f 0 = 0 Since, lim x → 0 f x = f 0 = 0 Hence, f x is continuous at x = 0 Now, if x ≠ 0 , then - x ≤ f x x ≤ x Thus,