Question

The value of f(0) so that the function \(f(x) = \frac{(256-8x)^{\frac{1}{4}}-4}{16-4(64+3x)^{\frac{1}{3}}}\), \(x\neq 0\) is continuous at \(x = 0\), is

• A. \(-\frac{1}{8}\)
• B. \(\frac{1}{8}\)
• C. \(\frac{1}{64}\)
• D. \(8\)

Hint:

Take the limit as x tends to 0 using binomial expansions, which equals the value of f(0).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For continuity at \(x = 0\), \(f(0)\) must equal \(\lim_{x\to0} f(x)\). Both numerator and denominator tend to 0, so we expand with the binomial series.

Step 2: Numerator
\((256 - 8x)^{1/4} = 4\left(1 - \frac{x}{32}\right)^{1/4} \approx 4\left(1 - \frac{x}{128}\right) = 4 - \frac{x}{32}\). So the numerator is \(\approx -\frac{x}{32}\).

Step 3: Denominator
\((64 + 3x)^{1/3} = 4\left(1 + \frac{3x}{64}\right)^{1/3} \approx 4\left(1 + \frac{x}{64}\right)\). So \(16 - 4(64 + 3x)^{1/3} \approx 16 - 16 - \frac{x}{4} = -\frac{x}{4}\).
\[ \lim_{x\to0} f(x) = \frac{-x/32}{-x/4} = \frac18 \]
So \(f(0) = \frac18\), option (B). Option (A) has the wrong sign because both terms are negative.

Final Answer:
\(f(0) = \frac18\), option (B). \[ \boxed{\frac{1}{8}} \]