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

Step 1: Understanding the Concept:
This is a limit problem that results in the indeterminate form \(\frac{0}{0}\) upon direct substitution. To solve it, we can use techniques like multiplying by the conjugate, L'Hôpital's Rule, or using trigonometric identities and series expansions. Multiplying by the conjugate is often the most direct method for limits involving square roots.
Step 2: Key Formula or Approach:
1. Multiply the numerator and denominator by the conjugate of the denominator, which is \((\sqrt{2} + \sqrt{1+\cos x})\). 2. Use the half-angle identity \(1 - \cos x = 2\sin^2(x/2)\). 3. Use the standard limit \(\lim_{u \to 0} \frac{\sin u}{u} = 1\).
Step 3: Detailed Explanation:
The limit is \(\lim_{x \to 0} \frac{x^2}{\sqrt{2} - \sqrt{1+\cos x}}\). Multiply by the conjugate: \[ \lim_{x \to 0} \frac{x^2}{\sqrt{2} - \sqrt{1+\cos x}} \times \frac{\sqrt{2} + \sqrt{1+\cos x}}{\sqrt{2} + \sqrt{1+\cos x}} \] The denominator becomes \((\sqrt{2})^2 - (\sqrt{1+\cos x})^2 = 2 - (1+\cos x) = 1 - \cos x\). The expression becomes: \[ \lim_{x \to 0} \frac{x^2 (\sqrt{2} + \sqrt{1+\cos x})}{1 - \cos x} \] Now, use the half-angle identity \(1 - \cos x = 2\sin^2(x/2)\): \[ \lim_{x \to 0} \frac{x^2 (\sqrt{2} + \sqrt{1+\cos x})}{2\sin^2(x/2)} \] We can split the limit. The term \((\sqrt{2} + \sqrt{1+\cos x})\) can be evaluated by direct substitution, as it's not causing the indeterminate form: As \(x \to 0\), \(\cos x \to 1\), so \(\sqrt{2} + \sqrt{1+\cos x} \to \sqrt{2} + \sqrt{1+1} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}\). So we have: \[ (2\sqrt{2}) \lim_{x \to 0} \frac{x^2}{2\sin^2(x/2)} = \sqrt{2} \lim_{x \to 0} \frac{x^2}{\sin^2(x/2)} \] To use the standard limit, we rewrite the expression: \[ \sqrt{2} \lim_{x \to 0} \frac{(x/2 \cdot 2)^2}{\sin^2(x/2)} = \sqrt{2} \lim_{x \to 0} \frac{4(x/2)^2}{\sin^2(x/2)} = 4\sqrt{2} \lim_{x \to 0} \left(\frac{x/2}{\sin(x/2)}\right)^2 \] Since \(\lim_{u \to 0} \frac{u}{\sin u} = 1\), the limit of the squared term is \(1^2 = 1\). The final result is: \[ 4\sqrt{2} \cdot 1 = 4\sqrt{2} \] Step 4: Final Answer:
The limit is equal to \(4\sqrt{2}\).