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: L Hopital route
Both parts vanish at 0, so differentiate top and bottom.

Step 2: Derivatives at 0
Numerator: $\frac14(256)^{-3/4}(-8) = \frac14 \cdot \frac{1}{64} \cdot (-8) = -\frac{1}{32}$. Denominator: $-4 \cdot \frac13 (64)^{-2/3} \cdot 3 = -4 \cdot \frac{1}{16} = -\frac14$.

Step 3: Ratio
$\dfrac{-1/32}{-1/4} = \dfrac18$. Option (B).

Final Answer:
Option (B). \[ \boxed{\frac{1}{8}} \]