The value of `f(0),` so that the function `f(x)=((27-2x)^(1/3)-3)/(9-3(243+5x)^(1//5))(x!=0)` is continuous, is given by (a)`2/3` (b) 6 (c) 2 (d) 4

Explanation

`f(x) = ((27- 2x)^(1/3) - 3)/(9- 3(243 + 5x)^(1/5))`

`= (3(1- 2x/27)^(1/3) - 3)/(9- 3*3*(1+ (5x)/(243))^(1/5))`

`= (3(1-(2x)/27 xx 1/3) -3)/(9- 9(1 + (5x)/243) xx 1/5)`

`= (3 - (2x)/27 - 3)/(9 - 9 -9 xx (5x)/243 xx 1/5)`

`= ((-2x)/27)/((-x)/27)`

`= 2`

option c is correct