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

To find the limit \(\lim_{x \to 2} \frac{2 - \sqrt{2 + x}}{2^{1/3} - (4 - x)^{1/3}}\), we begin by substituting \(x = 2\). Direct substitution results in an indeterminate form \(\frac{0}{0}\), which suggests the use of algebraic manipulation or L'Hôpital's Rule. However, algebraic manipulation is more straightforward here.

1. Rewrite the expression in a factorizable form:
2. For the numerator: \(\frac{2 - \sqrt{2 + x}}{2^{1/3} - (4-x)^{1/3}}\). 
   Let \(u = \sqrt{2 + x}\) and \(v = (4 - x)^{1/3}\).
3. As \(x \to 2\), \(u \to \sqrt{4} = 2\) and \(v \to (4 - 2)^{1/3} = 2^{1/3}\).
4. Use the expansions \((a^n - b^n) = (a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1})\) to rationalize.
5. Rationalize the numerator with the conjugate: 
   Multiply both numerator and denominator by \(2 + \sqrt{2 + x}\) to obtain: 
   \(\frac{(2 - \sqrt{2 + x})(2 + \sqrt{2 + x})}{(2^{1/3} - (4-x)^{1/3})(2 + \sqrt{2 + x})}\). 
   This becomes \(\frac{4 - (2+x)}{(2^{1/3} - (4-x)^{1/3})(2 + \sqrt{2 + x})}\).
6. Simplify to get the revised limit expression: 
   \(\frac{2 - x}{(2^{1/3} - (4-x)^{1/3})(2 + \sqrt{2 + x})}\).
7. Now address the denominator using the identity for cubes: 
   \((a^{1/3} - b^{1/3}) = \frac{a - b}{a^{2/3} + a^{1/3}b^{1/3} + b^{2/3}}\).
8. Recognize that as \(x\) approaches 2, the terms simplify significantly.
9. Substitute: \((2^{1/3} - (4-x)^{1/3}) \approx \frac{-x + 2}{ 3 \cdot 2^{4/3}}\).
10. Thus, the limit becomes: 
    \(\frac{1}{3 \cdot 2^{4/3} \cdot (2 + \sqrt{4})}\).
11. This simplifies to: \((-3 \cdot 2^{-4/3})\).

Hence, the correct answer is \(-3 \cdot 2^{-4/3}\).