Question

Test the continuity of the following function at the points indicated against them:

f(x) = `(x^3 - 27)/(x^2 - 9)`  for 0 ≤ x <3

      = `9/2`             for 3 ≤ x ≤ 6, at x = 3

Answer 1

f(3) = `9/2`  ...(given)

`lim_(x→3) "f"(x) = lim_(x→3) (x^3 - 27)/(x^2 - 9)`

= `lim_(x→3)((x - 3)(x^2 + 3x + 9))/((x - 3)(x + 3))`

= `lim_(x→3) (x^2 + 3x + 9)/(x + 3)   ...[("As"  x → 3","  x ≠ 3),(therefore x - 3 ≠ 0)]`

= `((3)^2 + 3(3) + 9)/(3 + 3)`

= `(9 + 9 + 9)/6`

= `27/6`

= `9/2`

∴ `lim_{x→3} "f"(x) = "f"(3)`
∴ Function f is continuous at x = 3