Key Concepts and Formulas

• Limit Evaluation Techniques: We will use algebraic manipulation, specifically multiplying by the conjugate, and the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
• Indeterminate Forms: Recognizing and resolving indeterminate forms like $\frac{0}{0}$ is crucial.
• Trigonometric Identities: The identity $\sin^2 x = 1 - \cos^2 x$ will be used.

Step-by-Step Solution

Part 1: Evaluating 'a'

• Step 1: Define the limit for 'a' and identify the indeterminate form. We are given $a=\lim\limits_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$. As $x \rightarrow 0$, the numerator approaches $\sqrt{1+\sqrt{1+0}} - \sqrt{2} = \sqrt{1+1} - \sqrt{2} = \sqrt{2} - \sqrt{2} = 0$. The denominator also approaches $0^4 = 0$. Thus, we have the indeterminate form $\frac{0}{0}$.

• Step 2: Multiply the numerator and denominator by the conjugate of the numerator. To resolve the $\frac{0}{0}$ form, we multiply by $\frac{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}$: $a = \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} \times \frac{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}$ $a = \lim _{x \rightarrow 0} \frac{(1+\sqrt{1+x^4}) - 2}{x^4 (\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})}$ $a = \lim _{x \rightarrow 0} \frac{\sqrt{1+x^4} - 1}{x^4 (\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})}$

• Step 3: Multiply the numerator and denominator by the conjugate of the new numerator. We again have an indeterminate form $\frac{0}{0}$ as $x \rightarrow 0$ in the expression $\frac{\sqrt{1+x^4} - 1}{x^4}$. We multiply by $\frac{\sqrt{1+x^4}+1}{\sqrt{1+x^4}+1}$: $a = \lim _{x \rightarrow 0} \frac{\sqrt{1+x^4} - 1}{x^4 (\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})} \times \frac{\sqrt{1+x^4}+1}{\sqrt{1+x^4}+1}$ $a = \lim _{x \rightarrow 0} \frac{(1+x^4) - 1}{x^4 (\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)}$ $a = \lim _{x \rightarrow 0} \frac{x^4}{x^4 (\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)}$

• Step 4: Cancel out the $x^4$ term and apply the limit. Since $x \rightarrow 0$ but $x \neq 0$, we can cancel $x^4$: $a = \lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)}$ Now, substitute $x=0$: $a = \frac{1}{(\sqrt{1+\sqrt{1+0}}+\sqrt{2})(\sqrt{1+0}+1)}$ $a = \frac{1}{(\sqrt{1+1}+\sqrt{2})(\sqrt{1}+1)}$ $a = \frac{1}{(\sqrt{2}+\sqrt{2})(1+1)}$ $a = \frac{1}{(2\sqrt{2})(2)}$ $a = \frac{1}{4\sqrt{2}}$

Part 2: Evaluating 'b'

• Step 5: Define the limit for 'b' and identify the indeterminate form. We are given $b=\lim\limits _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$. As $x \rightarrow 0$, $\sin^2 x \rightarrow \sin^2 0 = 0$. The denominator approaches $\sqrt{2}-\sqrt{1+\cos 0} = \sqrt{2}-\sqrt{1+1} = \sqrt{2}-\sqrt{2} = 0$. Thus, we have the indeterminate form $\frac{0}{0}$.

• Step 6: Use the trigonometric identity $\sin^2 x = 1 - \cos^2 x$. We can rewrite the numerator as $1 - \cos^2 x$: $b = \lim _{x \rightarrow 0} \frac{1-\cos^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$

• Step 7: Factor the numerator and multiply by the conjugate of the denominator. The numerator is a difference of squares: $1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$. To resolve the $\frac{0}{0}$ form in the denominator, we multiply by $\frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}$: $b = \lim _{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)}{\sqrt{2}-\sqrt{1+\cos x}} \times \frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}$ $b = \lim _{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{(\sqrt{2})^2 - (\sqrt{1+\cos x})^2}$ $b = \lim _{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{2 - (1+\cos x)}$ $b = \lim _{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{1 - \cos x}$

• Step 8: Cancel out the $(1-\cos x)$ term and apply the limit. Since $x \rightarrow 0$, $1-\cos x \rightarrow 0$, but for $x$ close to 0 and non-zero, $1-\cos x \neq 0$. We can cancel the term: $b = \lim _{x \rightarrow 0} (1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$ Now, substitute $x=0$: $b = (1+\cos 0)(\sqrt{2}+\sqrt{1+\cos 0})$ $b = (1+1)(\sqrt{2}+\sqrt{1+1})$ $b = (2)(\sqrt{2}+\sqrt{2})$ $b = 2(2\sqrt{2})$ $b = 4\sqrt{2}$

Part 3: Calculating $ab^3$

• Step 9: Substitute the values of 'a' and 'b' into the expression $ab^3$. We found $a = \frac{1}{4\sqrt{2}}$ and $b = 4\sqrt{2}$. $ab^3 = \left(\frac{1}{4\sqrt{2}}\right) \times (4\sqrt{2})^3$

• Step 10: Simplify the expression. $(4\sqrt{2})^3 = 4^3 \times (\sqrt{2})^3 = 64 \times (2\sqrt{2}) = 128\sqrt{2}$ So, $ab^3 = \frac{1}{4\sqrt{2}} \times 128\sqrt{2}$ $ab^3 = \frac{128\sqrt{2}}{4\sqrt{2}}$ $ab^3 = \frac{128}{4}$ $ab^3 = 32$

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with squaring and cubing terms, especially with radicals. Double-check each algebraic manipulation.
• Indeterminate Form Recognition: Always check for indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before applying limit evaluation techniques. If it's not indeterminate, the limit can often be found by direct substitution.
• Radical Conjugate: When dealing with limits involving square roots that result in $\frac{0}{0}$, multiplying by the conjugate is a standard and effective technique to simplify the expression.

Summary

To find the value of $ab^3$, we first evaluated the limit for 'a' by repeatedly multiplying by conjugates to resolve the indeterminate $\frac{0}{0}$ form. This yielded $a = \frac{1}{4\sqrt{2}}$. Next, we evaluated the limit for 'b' by using a trigonometric identity and then multiplying by the conjugate of the denominator to resolve the indeterminate $\frac{0}{0}$ form, resulting in $b = 4\sqrt{2}$. Finally, we substituted these values into the expression $ab^3$ and simplified to obtain the result.

The final answer is \boxed{32}.