Question

Discuss continuity of f(x) =`(x^3-64)/(sqrt(x^2+9)-5)` For x ≠ 4 

= 10 for x = 4  at x = 4

Answer 1

`f(4) = 10`

`lim_(x→4)f(x)=lim_(x→4)(x^3-64)/(sqrt(x^2+9)-5)`

`=lim_(x→4)(x^3-64)/(sqrt(x^2+9)-5)`

`lim_(x→4)(x^3-4^3)/(sqrt(x^2+9)-5)xx(sqrt(x^2+9)+5)/(sqrt(x^2+9)+5)`

`Lim_(x→4)((x^3-4^3)(sqrtx^2+9+5))/((sqrtx^2+9)^2-(-5)^2)`

`Lim_(x→4) ((x-4)(x^2+4x+16)(sqrt(x^2+9)+5))/(x^2+9-25)`    .......`[a^3 – b^3 = (a – b) (a^2 + ab + b^2))`

`Lim_(x→4) ((x-4)(x^2+4x+16)(sqrt(x^2+9)+5))/(x^2-16)`

`Lim_(x→4) ((x-4)(x^2+4x+16)(sqrt(x^2+9)+5))/((x-4)(x+4)`)

`Lim_(x→4) ((x^2+4x+16)(sqrt(x^2+9)+5))/(x+4)`

`(((4)^2+4(4)+16)(sqrt(4^2+9)+5))/(4+4)`

`((16+16+16)(sqrt(16+9)+5))/8`

`((16+16+16)(sqrt(25)+5))/8`

`((16+16+16)(5+5))/8`

`(48x10)/8`

`f(x)lim_(x→4)=60`

`lim_(x→4) f(x)≠f(4)` 

∴ f (x) is not continuous at x = 4.