Question

\( \lim_{x \to 0} \frac{x^2}{\sqrt{2} - \sqrt{1 + \cos x}} \) is equal to

• A. \( 4\sqrt{2} \)
• B. \(4 \)
• C. \( 2\sqrt{2} \)
• D. \( \sqrt{2} \)
• E. \(0 \)

Hint:

For limits, use standard approximations and rationalization to simplify radicals.

Answer 1

The Correct Option is A

Concept: Use approximation: \[ \cos x \approx 1 - \frac{x^2}{2} \]

Step 1: Simplify denominator.
\[ \sqrt{1+\cos x} \approx \sqrt{2 - \frac{x^2}{2}} \]

Step 2: Rationalize.
\[ \frac{x^2}{\sqrt{2} - \sqrt{1+\cos x}} \cdot \frac{\sqrt{2} + \sqrt{1+\cos x}}{\sqrt{2} + \sqrt{1+\cos x}} \] \[ = \frac{x^2(\sqrt{2} + \sqrt{1+\cos x})}{2 - (1+\cos x)} \]

Step 3: Simplify denominator.
\[ 2 - (1+\cos x) = 1 - \cos x \approx \frac{x^2}{2} \]

Step 4: Final simplification.
\[ = \frac{x^2(\sqrt{2} + \sqrt{2})}{x^2/2} = \frac{2x^2\sqrt{2}}{x^2/2} = 4\sqrt{2} \]