© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Limits to Evaluate
We need to find two limits:
$a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4}$
and
$b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}$.
Then compute $a \, b^3$.
Step 2: Evaluate $a$
Consider the numerator for $a$:
$\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}$.
For very small $x$, use the approximation:
$\sqrt{1 + x^4} \approx 1 + \frac{x^4}{2}$.
Hence,
$1 + \sqrt{1 + x^4} \approx 1 + \left(1 + \frac{x^4}{2}\right) = 2 + \frac{x^4}{2}.$
Next, for a small $\epsilon$, $\sqrt{2 + \epsilon} \approx \sqrt{2} + \frac{\epsilon}{2\sqrt{2}}.$
Here, $\epsilon = \frac{x^4}{2}$, so
$\sqrt{2 + \frac{x^4}{2}} \approx \sqrt{2} + \frac{\frac{x^4}{2}}{2\sqrt{2}}
= \sqrt{2} + \frac{x^4}{4\sqrt{2}}.$
Therefore,
$\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}
\approx \left(\sqrt{2} + \frac{x^4}{4\sqrt{2}}\right) - \sqrt{2}
= \frac{x^4}{4\sqrt{2}}.$
Dividing by $x^4$,
$\frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4}
\approx \frac{\frac{x^4}{4\sqrt{2}}}{x^4}
= \frac{1}{4\sqrt{2}}.$
Thus,
$a = \frac{1}{4\sqrt{2}}.$
Step 3: Evaluate $b$
Now, consider
$b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}.$
For small $x$, $\sin x \approx x$, so $\sin^2 x \approx x^2.$
Also, $\cos x \approx 1 - \frac{x^2}{2}.$
Thus,
$1 + \cos x \approx 2 - \frac{x^2}{2}.$
Next, use the approximation for small $\epsilon$: $\sqrt{2 - \epsilon} \approx \sqrt{2} - \frac{\epsilon}{2\sqrt{2}}.$
Here, $\epsilon = \frac{x^2}{2}$, so
$\sqrt{2 - \frac{x^2}{2}} \approx \sqrt{2} - \frac{\frac{x^2}{2}}{2\sqrt{2}}
= \sqrt{2} - \frac{x^2}{4\sqrt{2}}.$
Hence,
$\sqrt{2} - \sqrt{1 + \cos x}
\approx \sqrt{2} - \left(\sqrt{2} - \frac{x^2}{4\sqrt{2}}\right)
= \frac{x^2}{4\sqrt{2}}.$
Therefore,
$b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}
\approx \lim_{x \to 0} \frac{x^2}{\frac{x^2}{4\sqrt{2}}}
= 4\sqrt{2}.$
Step 4: Calculate $a \, b^3$
We have $a = \frac{1}{4\sqrt{2}}$ and $b = 4\sqrt{2}.$
First compute $b^3$:
$(4\sqrt{2})^3 = 4^3 \times (\sqrt{2})^3 = 64 \times 2\sqrt{2} = 128\sqrt{2}.$
Hence,
$a \, b^3 = \frac{1}{4\sqrt{2}} \times 128\sqrt{2}
= \frac{128\sqrt{2}}{4\sqrt{2}}
= 32.$
Conclusion
Therefore, the value of $a \, b^3$ is $32.$
