Let f : [ 1 , 3 ] → R be a function satisfying [ x ] x ≤ f ( x ) ≤ 6 − x , for all x  = 2 and f ( 2 ) = 1 , where R is the set of all real numbers and [ x ] denotes the largest integer less than

Question

Let f : [ 1 , 3 ] → R be a function satisfying [ x ] x ​ ≤ f ( x ) ≤ 6 − x ​ , for all x  = 2 and f ( 2 ) = 1 , where R is the set of all real numbers and [ x ] denotes the largest integer less than or equal to x . Statement 1: lim x → 2 − ​ f ( x ) exists. Statement 2: f is continuous at x = 2 .

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​ Consider [ x ] x ​ ≤ f ( x ) ≤ 6 − x ​ ⇒ x → 2 − lim ​ [ x ] x ​ = 1 2 ​ = 2 ​ ​ ⇒ x → 2 − lim ​ 6 − x ​ = 2 ∴ x → 2 − lim ​ f ( x ) = 2 ​ [By Sandwich theorem] Now lim x → 2 + ​ [ x ] x ​ = 1 , lim x → 2 + ​ 6 − x ​ = 2 Hence by Sandwich theorem lim x → 2 + ​ f ( x ) does not exists. Therefore f is not continuous at x = 2 . Thus statement-1 is true but statement- 2 is not true

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