If $a=\underset{x\to <!--\; \to \; -->0}{lim}\frac{\sqrt{1+\sqrt{1+x^4}}-<!--\; -\; -->\sqrt{2}}{x^4}$ and $b=\underset{x\to <!--\; \to \; -->0}{lim}\frac{{\sin }^2<!--\; \; -->x}{\sqrt{2}-<!--\; -\; -->\sqrt{1+\cos <!--\; \; -->x}}$ , then the value of ab³ is :

The correct option is 32.

To find the value of $a{b}^3$, we need to evaluate the limits $a$ and $b$ step-by-step.

Step 1: Find the value of $a$
The limit $a$ is given by:
$$a=\underset{x\to 0}{lim}\frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$$
Let $y=x^4$. As $x\to 0$, we have $y\to 0$. Substituting this, we get:
$$a=\underset{y\to 0}{lim}\frac{\sqrt{1+\sqrt{1+y}}-\sqrt{2}}{y}$$
To evaluate this limit, we rationalize the numerator:
$$a=\underset{y\to 0}{lim}\frac{(\sqrt{1+\sqrt{1+y}}-\sqrt{2})(\sqrt{1+\sqrt{1+y}}+\sqrt{2})}{y(\sqrt{1+\sqrt{1+y}}+\sqrt{2})}$$
Using the algebraic identity $(A-B)(A+B)=A^2-B^2$, the numerator simplifies to:
$$(1+\sqrt{1+y})-2=\sqrt{1+y}-\mathrm{1<!--\; -->}$$
Thus, the limit expression becomes:
$$a=\underset{y\to 0}{lim}\frac{\sqrt{1+y}-1}{y(\sqrt{1+\sqrt{1+y}}+\sqrt{2})}$$
Now, rationalizing the numerator once more:
$$a=\underset{y\to 0}{lim}\frac{(\sqrt{1+y}-1)(\sqrt{1+y}+1)}{y(\sqrt{1+\sqrt{\mathrm{1<!--\; --><mo>+</mo>\; <mi>y</mi>}}}+\sqrt{2})(\sqrt{1+y}+1)}$$
Simplifying the numerator gives:
$$(1+y)-\mathrm{1<!--\; <mo-->=\; <mi>y</mi>}$$
Substituting this back, we can cancel $y$ from both the numerator and the denominator:
$$a=\underset{y\to 0}{lim}\frac{y}{y(\sqrt{1+\sqrt{1+y}}+\sqrt{2})(\sqrt{1+y}+1)}$$

Evaluating this limit directly by setting $y=0$:
$$a=\frac{1}{(\sqrt{1+\sqrt{1+0}}+\sqrt{2})(\sqrt{1+0}+1)}$$

Step 2: Find the value of $b$
The limit $b$ is given by:
$$b=\underset{x\to 0}{lim}\frac{{sin}^{2}x}{\sqrt{2}-\sqrt{1+cosx}}$$
First, rationalize the denominator:
$$b=\underset{x\to 0}{lim}\frac{{sin}^{2}x(\sqrt{2}+\sqrt{1+cosx})}{(\sqrt{2}-\sqrt{1+cosx})(\sqrt{2}+\sqrt{1+cosx})}$$
The denominator simplifies to:

Thus:
$$b=\underset{x\to 0}{lim}\frac{{sin}^{2}x(\sqrt{2}+\sqrt{1+cosx})}{1-cosx}$$
We know the trigonometric identity ${sin}^{2}x=1-{cos}^{2}x=(1-cosx)(1+cosx)$. Substituting this in:
$$b=\underset{x\to 0}{lim}\frac{(1-cosx)(\mathrm{1<!--\; <mo-->+\; <mo>cos</mo>\; <mi>x</mi>\; <mo>)</mo>}(\sqrt{2}+\sqrt{1+cosx})}{\mathrm{1<!--\; <mo-->-\; <mo>cos</mo>\; <mi>x</mi>}}$$
Canceling out the common term $(1-cosx)$ (since $x\ne 0$ near the limit):
$$b=\underset{x\to 0}{lim}(\mathrm{1<!--\; <mo-->+\; <mo>cos</mo>\; <mi>x</mi>\; <mo>)</mo>}(\sqrt{2}+\sqrt{1+cosx})$$
Evaluating the limit by substituting $x=0$:

Since $cos0=1$:

Step 3: Calculate the value of $a{b}^3$
Using our values $a=\frac{1}{4\sqrt{2}}$ and $b=4\sqrt{2}$:
$$a{b}^3=\left(\frac{1}{4\sqrt{2}}\right)\times {\left(4\sqrt{2}\right)}^3$$
$$a{b}^3={\left(4\sqrt{2}\right)}^2$$
$$a{b}^3=16\times 2=32$$

Thus, the value of $a{b}^3$ is 32.