Question

\(\lim_{x \to 2} \frac{2 - \sqrt{2 + x}}{2^{1/3} - (4 - x)^{1/3}}\) is equal to

• A. \(2\cdot 3^{-1/2}\)
• B. \(3\cdot 2^{-4/3}\)
• C. \(-3\cdot 2^{-4/3}\)
• D. None of these

Hint:

For roots, use expansion: \((1+h)^n \approx 1+nh\) for small \(h\).

Answer 1

The Correct Option is C

Concept: Use substitution and series/expansion for limits of type \(0/0\).

Step 1: Substitute \(x=2+h\).
\[ \text{Numerator: } 2 - \sqrt{4+h} \approx 2 - \left(2 + \frac{h}{4}\right) = -\frac{h}{4} \]

Step 2: Denominator.
\[ (4-x)^{1/3} = (2-h)^{1/3} \approx 2^{1/3}\left(1 - \frac{h}{2}\right)^{1/3} \] \[ \approx 2^{1/3}\left(1 - \frac{h}{6}\right) = 2^{1/3} - \frac{2^{1/3}h}{6} \] \[ \Rightarrow 2^{1/3} - (4-x)^{1/3} \approx \frac{2^{1/3}h}{6} \]

Step 3: Take ratio.
\[ \frac{-\frac{h}{4}}{\frac{2^{1/3}h}{6}} = -\frac{6}{4\cdot 2^{1/3}} = -\frac{3}{2\cdot 2^{1/3}} = -3\cdot 2^{-4/3} \] Conclusion: \[ {(C)} \]