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 answer is 32.

We need to evaluate two limits, $a$ and $b$, then compute $a{b}^3$.

Step 1: Evaluate $a$

a=\underset{x\to 0}{lim}\frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}

As $x\to 0$, the numerator and denominator both go to 0, giving the 0/0 indeterminate form. We rationalize the numerator by multiplying and dividing by the conjugate $\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}$:

a=\underset{x\to 0}{lim}\frac{(1+\sqrt{1+x^4})-2}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})}=\underset{x\to 0}{lim}\frac{\sqrt{1+x^4}-1}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})}

Rationalize the numerator again by multiplying by $\sqrt{1+x^4}+1$:

a=\underset{x\to 0}{lim}\frac{(1+x^4)-1}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)}=\underset{x\to 0}{lim}\frac{x^4}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)}

Cancel $x^4$ and substitute $x=0$:

a=\frac{1}{(\sqrt{2}+\sqrt{2})(1+1)}=\frac{1}{2\sqrt{2}·2}=\frac{1}{4\sqrt{2}}

Step 2: Evaluate $b$

b=\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}

Again a 0/0 form. Rationalize the denominator by multiplying and dividing by $\sqrt{2}+\sqrt{1+\cos x}$:

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

Now use the key identity ${\sin }^{2}x=1-{\cos }^{2}x=(1-\cos x)(1+\cos x)$:

b=\underset{x\to 0}{lim}\frac{(1-\cos x)(1+\cos x)·(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}

Cancel $(1-\cos x)$ and substitute $x=0$ (so $\cos x=1$):

b=(1+1)·(\sqrt{2}+\sqrt{1+1})=2·(\sqrt{2}+\sqrt{2})=2·2\sqrt{2}=4\sqrt{2}

Step 3: Compute $a{b}^3$

First compute $b^3$:

b^3={\left(4\sqrt{2}\right)}^3=64·{\left(\sqrt{2}\right)}^3=64·2\sqrt{2}=128\sqrt{2}

Now multiply by $a=\frac{1}{4\sqrt{2}}$:

a{b}^3=\frac{1}{4\sqrt{2}}·128\sqrt{2}=\frac{128\sqrt{2}}{4\sqrt{2}}=\frac{128}{4}=32

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