Test the continuity of the following function at the points 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→) ((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 ...[(because x→ 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