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Question
Test the continuity of the following function at the point or interval indicated against them :
f(x) `{:(= (x^3 - 8)/(sqrt(x + 2) - sqrt(3x - 2))",", "for" x ≠ 2),(= -24",", "for" x = 2):}}` at x = 2
Solution
f(2) = – 24 …(given)
`lim_(x -> 2) "f"(x) = lim_(x -> 2) (x^3 - 8)/(sqrt(x + 2) - sqrt(3x - 2))`
= `lim_(x -> 2) (x^3 - 8)/(sqrt(x + 2) - sqrt(3x - 2)) xx (sqrt(x + 2) + sqrt(3x + 2))/(sqrt(x + 2) + sqrt(3x - 2))`
= `lim_(x -> 2) ((x^3 - 8)(sqrt(x + 2) + sqrt(3x - 2)))/((x + 2) - (3x - 2))`
= `lim_(x -> 2) ((x^3 - 2^3)(sqrt(x + 2) + sqrt(3x - 2)))/(-2x + 4)`
= `lim_(x -> 2) ((x - 2)(x^2 + 2x + 4)(sqrt(x + 2) + sqrt(3x - 2)))/(-2(x - 2))`
= `lim_(x -> 2) ((x^2 + 2x + 4)(sqrt(x + 2) + sqrt(3x - 2)))/(-2) ...[(becausex -> 2"," x ≠ 2),(therefore x - 2 ≠ 0)]`
= `(-1)/(2) lim_(x -> 2) (x^2 + 2x + 4) (sqrt(x + 2) + sqrt(3x - 2))`
= `(-1)/(2) lim_(x -> 2) (x^2 + 2x + 4) lim_(x -> 2) (sqrt(x + 2) + sqrt(3x - 2))`
= `(-1)/(2) xx [2^2 + 2(2) + 4] xx (sqrt(2 + 2) + sqrt(3(2) - 2))`
= `(-1)/(2) xx 12 xx (2 + 2)`
= – 24
∴ `lim_(x -> 2) "f"(x)` = f(2)
∴ f(x) is continuous at x = 2
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