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

2

Since f (x) s continuous at x = 0, therefore

$f(0)=\underset{x→0}{\lim}f(x)=\underset{x→0}{\lim}\frac{(27-2x)^{1/3}-3}{9-3(243+5x)^{1/5}}$  $(form\,\frac{0}{0})$

$=\underset{x→0}{\lim}\frac{\frac{1}{3}(27-2x)^{-2/3}(-2)}{-\frac{3}{5}(243+5x)^{-4/5}(5)}=2$