If $\mathrm{f}(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}, x \neq 0$ is continuous at $x=0$, then the value of $f(0)$ is

To ensure continuity at $x=0$, define $f(0) = \lim_{x \to 0} f(x)$ where

$f(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}$.

Direct substitution yields the indeterminate form $\frac{0}{0}$, requiring further analysis.

Applying L'Hôpital's rule, differentiate numerator and denominator:

$N'(x) = -\frac{2}{3}(27-2x)^{-\frac{2}{3}}$

$D'(x) = -3(243+5x)^{-\frac{4}{5}}$

The limit becomes

$\lim_{x \to 0} \frac{N'(x)}{D'(x)} = \frac{-\frac{2}{3}(27)^{-\frac{2}{3}}}{-3(243)^{-\frac{4}{5}}}$.

Expressing $27=3^3$ and $243=3^5$, we simplify:

$\frac{-\frac{2}{3} \cdot 3^{-2}}{-3 \cdot 3^{-4}} = \frac{-\frac{2}{27}}{-\frac{1}{27}} = 2$.

Alternatively, using the standard limit $\lim_{y \to 0} \frac{(a+y)^n - a^n}{y} = n a^{n-1}$, we compute numerator and denominator limits separately.

The numerator limit transforms to $-2 \cdot \frac{1}{3}(27)^{-\frac{2}{3}} = -\frac{2}{27}$.

The denominator limit becomes $-15 \cdot \frac{1}{5}(243)^{-\frac{4}{5}} = -\frac{1}{27}$.

The ratio confirms $\lim_{x \to 0} f(x) = 2$.

Final answer: $f(0) = 2$