Question

If \( a = \lim_{{x \to 0}} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4} \) and \( b = \lim_{{x \to 0}} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}} \), then the value of \( ab^3 \) is:

• A. 36
• B. 32
• C. 25
• D. 30

Answer 1

The Correct Option is B

Given:
\(a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4}\)
and

\(b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}\)

We need to find the value of \( a \cdot b^2 \).

Step 1. Finding \( a \): Consider:
 \(a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4}\)
  Rationalizing the numerator:

 \(a = \lim_{x \to 0} \frac{\left( \sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2} \right) \left( \sqrt{1 + \sqrt{1 + x^4}} + \sqrt{2} \right)}{x^4 \left( \sqrt{1 + \sqrt{1 + x^4}} + \sqrt{2} \right)}\)

  This gives:

 \(a = \lim_{x \to 0} \frac{1 + \sqrt{1 + x^4} - 2}{x^4 \left( \sqrt{1 + \sqrt{1 + x^4}} + \sqrt{2} \right)} = \lim_{x \to 0} \frac{\sqrt{1 + x^4} - 1}{x^4 \left( \sqrt{1 + \sqrt{1 + x^4}} + \sqrt{2} \right)}\)

  Approximating \( \sqrt{1 + x^4} \approx 1 + \frac{x^4}{2} \) as \( x \to 0 \):

  \(a = \lim_{x \to 0} \frac{\frac{x^4}{2}}{x^4 \left( \sqrt{1 + \sqrt{1 + x^4}} + \sqrt{2} \right)} = \frac{1}{4\sqrt{2}}\)

Step 2. Finding \( b \): Consider:
\(b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}\)
  Rationalizing the denominator:
\(b = \lim_{x \to 0} \frac{\sin^2 x \left( \sqrt{2} + \sqrt{1 + \cos x} \right)}{2 - (1 + \cos x)}\)
  Simplifying:

 \(b = \lim_{x \to 0} \frac{\sin^2 x \left( \sqrt{2} + \sqrt{1 + \cos x} \right)}{1 - \cos x}\)

  Using \( \sin^2 x = 1 - \cos^2 x \) and \( \lim_{x \to 0} \cos x = 1 \):

  \(b = 2 \left( \sqrt{2 + \sqrt{1 + \cos x}} \right) = 2 (\sqrt{2 + \sqrt{2}}) = 4\sqrt{2}\)

Step 3. Calculating \( a \cdot b^2 \):
  
  \(a \cdot b^2 = \frac{1}{4\sqrt{2}} \cdot (4\sqrt{2})^2 = 32\)

Answer: \((2) \, 32\)