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

Answer: (B)

32

Answer 2

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: $(2) \, 32$