Question

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

• A. $\frac{2}{3}$
• B. 6
• C. 2
• D. $\frac{1}{3}$

Hint:

Use L'Hôpital's Rule for complex radical limits in $0/0$ form.

Answer 1

The Correct Option is C

Step 1: Concept
For a function to be continuous at $x=0$, $f(0) = \lim_{x \to 0} f(x)$.

Step 2: Meaning
The limit is in $0/0$ form, so we can apply L'Hôpital's Rule or standard limits.

Step 3: Analysis
Differentiating numerator: $\frac{1}{3}(27-2x)^{-2/3}(-2)$. At $x=0$, this is $\frac{-2}{3(27)^{2/3}} = \frac{-2}{27}$. Differentiating denominator: $-3 \cdot \frac{1}{5}(243+5x)^{-4/5}(5) = -3(243+5x)^{-4/5}$. At $x=0$, this is $-3(243)^{-4/5} = -3(3^5)^{-4/5} = -3/81 = -1/27$.

Step 4: Conclusion
Limit $= \frac{-2/27}{-1/27} = 2$. Therefore, $f(0) = 2$. Final Answer: (C)