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प्रश्न
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
उत्तर
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
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