f(0) का मान क्या है, यदि `f(x)=((27-2x)^(1//3)-3)/(9-3(243+5x)^(1//5) (x ne 0),x=0` सतत है

f(x) = {{:(((27- 2x )^(1//3)-3)/(9-3 (243 + 5x)^(1//5)) " if " x ne 0),(k " if " x = 0 ):} continuous at x = 0 then k =

The value of f(0) so that the function f(x) = ((27-2x)^(1/3)-3)/(9-3(243+5x)^(1/5)), x ne 0 is continuous, is given by

The value of f(0) , so that the function f(x)=((27-2x)^(1/3)-3)/(9-3(243+5x)^(1/5)),(xne0) is continuous at x=0 , is given by

If f(x) is continuous at x=0 , where f(x)=((27-2x)^(1/3)-3)/(9-3(243+5x)^(1/5)) , for x!=0 then f(0)=

The value of f(0) so that the function f(x)=((27-2x)^(1//3)-3)/(9-3(243+5x)^(1//5))" is continuous at x=0 is "

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