Key Concepts and Formulas

• Factorization: Recognizing algebraic patterns to simplify expressions. Specifically, the expression inside the limit can be factored.
• Standard Limit of Cosine: The fundamental limit involving cosine is $\mathop {\lim }\limits_{y \to 0} \frac{1 - \cos y}{y^2} = \frac{1}{2}$. This is derived from the Taylor series expansion of $\cos y$ or using L'Hopital's rule.
• Taylor Series Expansion of Cosine: For small $z$, $\cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \dots$. This can be used to evaluate limits when the standard form is not directly applicable.
• Limit Properties: The limit of a product is the product of the limits, provided the individual limits exist. $\mathop {\lim }\limits_{x \to a} [f(x)g(x)] = \mathop {\lim }\limits_{x \to a} f(x) \cdot \mathop {\lim }\limits_{x \to a} g(x)$.

Step-by-Step Solution

Step 1: Simplify the Expression Inside the Limit The expression inside the limit is $1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2}\cos \frac{x^2}{4}$. We can factor this expression by grouping terms. $1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2}\cos \frac{x^2}{4} = \left(1 - \cos \frac{x^2}{2}\right) - \cos \frac{x^2}{4}\left(1 - \cos \frac{x^2}{2}\right)$ $= \left(1 - \cos \frac{x^2}{2}\right)\left(1 - \cos \frac{x^2}{4}\right)$ So the limit becomes: $\mathop {\lim }\limits_{x \to 0} \frac{1}{x^8} \left(1 - \cos \frac{x^2}{2}\right)\left(1 - \cos \frac{x^2}{4}\right)$

Step 2: Rearrange the Expression to Utilize the Standard Cosine Limit We want to manipulate the expression to match the form $\frac{1 - \cos y}{y^2}$. We have $x^8$ in the denominator, and the arguments of cosine are $\frac{x^2}{2}$ and $\frac{x^2}{4}$. We can rewrite $x^8$ as $\left(\frac{x^2}{2}\right)^2 \cdot 4 \cdot \left(\frac{x^2}{4}\right)^2 \cdot 16$. $\frac{1}{x^8} \left(1 - \cos \frac{x^2}{2}\right)\left(1 - \cos \frac{x^2}{4}\right) = \frac{\left(1 - \cos \frac{x^2}{2}\right)}{\left(\frac{x^2}{2}\right)^2} \cdot \frac{\left(1 - \cos \frac{x^2}{4}\right)}{\left(\frac{x^2}{4}\right)^2} \cdot \frac{\left(\frac{x^2}{2}\right)^2 \left(\frac{x^2}{4}\right)^2}{x^8}$ Let's simplify the last fraction: $\frac{\left(\frac{x^2}{2}\right)^2 \left(\frac{x^2}{4}\right)^2}{x^8} = \frac{\frac{x^4}{4} \cdot \frac{x^4}{16}}{x^8} = \frac{\frac{x^8}{64}}{x^8} = \frac{1}{64}$ So the expression becomes: $\frac{1}{64} \cdot \frac{\left(1 - \cos \frac{x^2}{2}\right)}{\left(\frac{x^2}{2}\right)^2} \cdot \frac{\left(1 - \cos \frac{x^2}{4}\right)}{\left(\frac{x^2}{4}\right)^2}$

Step 3: Evaluate the Limit Using the Standard Cosine Limit Formula Now we can apply the limit to each part of the expression. Let $y_1 = \frac{x^2}{2}$. As $x \to 0$, $y_1 \to 0$. $\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos \frac{x^2}{2}}{\left(\frac{x^2}{2}\right)^2} = \mathop {\lim }\limits_{y_1 \to 0} \frac{1 - \cos y_1}{y_1^2} \cdot \frac{1}{1} = \frac{1}{2}$ Let $y_2 = \frac{x^2}{4}$. As $x \to 0$, $y_2 \to 0$. $\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos \frac{x^2}{4}}{\left(\frac{x^2}{4}\right)^2} = \mathop {\lim }\limits_{y_2 \to 0} \frac{1 - \cos y_2}{y_2^2} \cdot \frac{1}{1} = \frac{1}{2}$ Now, applying the limit to the entire expression: $\mathop {\lim }\limits_{x \to 0} \left\{ \frac{1}{x^8}\left(1 - \cos \frac{x^2}{2}\right)\left(1 - \cos \frac{x^2}{4}\right) \right\} = \frac{1}{64} \cdot \left(\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos \frac{x^2}{2}}{\left(\frac{x^2}{2}\right)^2}\right) \cdot \left(\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos \frac{x^2}{4}}{\left(\frac{x^2}{4}\right)^2}\right)$ $= \frac{1}{64} \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{64} \cdot \frac{1}{4} = \frac{1}{256}$

Step 4: Equate the Result with the Given Expression and Solve for k The problem states that the limit is equal to $2^{-k}$. $\frac{1}{256} = 2^{-k}$ We know that $256 = 2^8$. $\frac{1}{2^8} = 2^{-k}$ $2^{-8} = 2^{-k}$ Equating the exponents: $-8 = -k$ $k = 8$

Common Mistakes & Tips

• Incorrect Factorization: Ensure the algebraic factorization of the cosine terms is done correctly. A common mistake is to misapply signs during grouping.
• Misapplication of Standard Limit: The standard limit is $\frac{1-\cos y}{y^2}$, not $\frac{1-\cos y}{y}$ or other variations. Make sure the denominator matches the square of the argument of cosine.
• Ignoring Constants: When rearranging terms to fit the standard limit, be careful not to drop or incorrectly calculate the constant factors that arise. For example, $\left(\frac{x^2}{2}\right)^2 = \frac{x^4}{4}$, not $x^4$.

Summary The problem requires evaluating a limit of a trigonometric expression. The first step involves factoring the expression within the limit. Subsequently, the factored expression is rearranged to utilize the standard limit $\mathop {\lim }\limits_{y \to 0} \frac{1 - \cos y}{y^2} = \frac{1}{2}$. By carefully applying this standard limit to the appropriate terms and managing the constant factors, the value of the limit is found to be $\frac{1}{256}$. Equating this result to $2^{-k}$ and solving for $k$ yields $k=8$.

The final answer is $\boxed{8}$.