If $\mathrm{a}=\lim\limits_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $\mathrm{b}=\lim\limits _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $a b^3$ is :

<p>$$\begin{aligned} a= & \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)} \\ & =\lim _{x \rightarrow 0} \frac{x^4}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)\left(\sqrt{1+x^4}+1\right)} \end{aligned}$$</p> <p>Applying limit $\mathrm{a}=\frac{1}{4 \sqrt{2}}$</p> <p>$$\begin{aligned} & b=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}} \\ & =\lim _{x \rightarrow 0} \frac{\left(1-\cos ^2 x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} \\ & b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x}) \end{aligned}$$</p> <p>Applying limits $b=2(\sqrt{2}+\sqrt{2})=4 \sqrt{2}$</p> <p>Now, $a b^3=\frac{1}{4 \sqrt{2}} \times(4 \sqrt{2})^3=32$</p>