Evaluate the limit a = _x o 0 sqrt(1+1+x4)-sqrt(2)x4

We have the expression for $a$:

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

This limit is of the indeterminate form $\frac{0}{0}$. We multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}$:

$$a = \lim_{x \to 0} \frac{\left(\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}\right)\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)}$$ $$a = \lim_{x \to 0} \frac{(1+\sqrt{1+x^4}) - 2}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)}$$ $$a = \lim_{x \to 0} \frac{\sqrt{1+x^4} - 1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)}$$

This is still of the form $\frac{0}{0}$. We multiply the numerator and denominator by the conjugate of $\sqrt{1+x^4}-1$, which is $\sqrt{1+x^4}+1$:

$$a = \lim_{x \to 0} \frac{(\sqrt{1+x^4} - 1)(\sqrt{1+x^4} + 1)}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)(\sqrt{1+x^4}+1)}$$ $$a = \lim_{x \to 0} \frac{(1+x^4) - 1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)(\sqrt{1+x^4}+1)}$$ $$a = \lim_{x \to 0} \frac{x^4}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)(\sqrt{1+x^4}+1)}$$

Cancel $x^4$ from the numerator and denominator (since $x \to 0$, $x \neq 0$):

$$a = \lim_{x \to 0} \frac{1}{\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)(\sqrt{1+x^4}+1)}$$

Now, substitute $x=0$:

$$a = \frac{1}{\left(\sqrt{1+\sqrt{1+0^4}}+\sqrt{2}\right)(\sqrt{1+0^4}+1)}$$ $$a = \frac{1}{\left(\sqrt{1+\sqrt{1}}+\sqrt{2}\right)(\sqrt{1}+1)}$$ $$a = \frac{1}{\left(\sqrt{1+1}+\sqrt{2}\right)(1+1)}$$ $$a = \frac{1}{\left(\sqrt{2}+\sqrt{2}\right)(2)}$$ $$a = \frac{1}{(2\sqrt{2})(2)}$$ $$a = \frac{1}{4\sqrt{2}}$$

Rationalize the denominator:

$$a = \frac{1}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{8}$$

We have the expression for $b$:

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

This limit is of the indeterminate form $\frac{0}{0}$. We multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2}+\sqrt{1+\cos x}$:

$$b = \lim_{x \to 0} \frac{\sin^2 x (\sqrt{2}+\sqrt{1+\cos x})}{(\sqrt{2}-\sqrt{1+\cos x})(\sqrt{2}+\sqrt{1+\cos x})}$$ $$b = \lim_{x \to 0} \frac{\sin^2 x (\sqrt{2}+\sqrt{1+\cos x})}{2 - (1+\cos x)}$$ $$b = \lim_{x \to 0} \frac{\sin^2 x (\sqrt{2}+\sqrt{1+\cos x})}{1 - \cos x}$$

Use the trigonometric identity $\sin^2 x = 1 - \cos^2 x = (1-\cos x)(1+\cos x)$:

$$b = \lim_{x \to 0} \frac{(1-\cos x)(1+\cos x) (\sqrt{2}+\sqrt{1+\cos x})}{1 - \cos x}$$

Cancel $(1-\cos x)$ from the numerator and denominator (since $x \to 0$, $1-\cos x \neq 0$ for $x$ near 0 but not equal to 0):

$$b = \lim_{x \to 0} (1+\cos x) (\sqrt{2}+\sqrt{1+\cos x})$$

Now, substitute $x=0$:

$$b = (1+\cos 0) (\sqrt{2}+\sqrt{1+\cos 0})$$ $$b = (1+1) (\sqrt{2}+\sqrt{1+1})$$ $$b = (2) (\sqrt{2}+\sqrt{2})$$ $$b = (2) (2\sqrt{2})$$ $$b = 4\sqrt{2}$$

Substitute the values of $a$ and $b$ we found:

$$ab^3 = \left(\frac{\sqrt{2}}{8}\right) (4\sqrt{2})^3$$ $$ab^3 = \left(\frac{\sqrt{2}}{8}\right) (4^3 \cdot (\sqrt{2})^3)$$ $$ab^3 = \left(\frac{\sqrt{2}}{8}\right) (64 \cdot 2\sqrt{2})$$ $$ab^3 = \left(\frac{\sqrt{2}}{8}\right) (128\sqrt{2})$$ $$ab^3 = \frac{\sqrt{2} \cdot 128\sqrt{2}}{8}$$ $$ab^3 = \frac{128 \cdot (\sqrt{2})^2}{8}$$ $$ab^3 = \frac{128 \cdot 2}{8}$$ $$ab^3 = \frac{256}{8}$$ $$ab^3 = 32$$

The final answer is 32.