Step 1: Understanding the Concept:
For a function to be continuous at $x = a$, the value of the function at that point must be equal to its limit as $x$ approaches $a$. Thus, $f(0) = \lim_{x \to 0} f(x)$.
Step 2: Key Formula or Approach:
The limit is in the indeterminate form $\frac{0}{0}$. We can use L'Hôpital's rule or binomial expansion for $(1+x)^n \approx 1+nx$ when $x$ is very small.
Step 3: Detailed Explanation:
Let's use the binomial expansion method as it's often cleaner for fractional powers.
First, rewrite the expression to apply the approximation $(1+y)^n \approx 1+ny$:
Numerator:
\[ (27 - 2x)^{1/3} - 3 = \left(27\left(1 - \frac{2x}{27}\right)\right)^{1/3} - 3 = 3\left(1 - \frac{2x}{27}\right)^{1/3} - 3 \]
For $x \to 0$, apply binomial expansion:
\[ \approx 3\left(1 + \frac{1}{3}\left(-\frac{2x}{27}\right)\right) - 3 = 3\left(1 - \frac{2x}{81}\right) - 3 = 3 - \frac{2x}{27} - 3 = -\frac{2x}{27} \]
Denominator:
\[ 9 - 3(243 + 5x)^{1/5} = 9 - 3\left(243\left(1 + \frac{5x}{243}\right)\right)^{1/5} = 9 - 3 \cdot 3\left(1 + \frac{5x}{243}\right)^{1/5} \]
\[ = 9 - 9\left(1 + \frac{5x}{243}\right)^{1/5} \]
Apply binomial expansion:
\[ \approx 9 - 9\left(1 + \frac{1}{5}\left(\frac{5x}{243}\right)\right) = 9 - 9\left(1 + \frac{x}{243}\right) = 9 - 9 - \frac{9x}{243} = -\frac{9x}{243} = -\frac{x}{27} \]
Now evaluate the limit by substituting the approximations back into the fraction:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{-\frac{2x}{27}}{-\frac{x}{27}} = \lim_{x \to 0} \left(-\frac{2x}{27} \times -\frac{27}{x}\right) = 2 \]
Since the function is continuous at $x=0$, $f(0) = \lim_{x \to 0} f(x) = 2$.
Step 4: Final Answer:
The value of $f(0)$ is 2.