Key Concepts and Formulas

• Standard Limit: The fundamental trigonometric limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
• Double Angle Identity: The trigonometric identity $\sin^2 \alpha = \frac{1 - \cos(2\alpha)}{2}$.
• Limit Properties: The limit of a product is the product of the limits, and the limit of a sum/difference is the sum/difference of the limits, provided the individual limits exist.
• Algebraic Manipulation: Using algebraic techniques to simplify expressions and make them amenable to standard limit forms.

Step-by-Step Solution

Step 1: Apply the double angle identity for $\sin^2 \alpha$. The expression involves $\sin^2(\dots)$. We can use the identity $\sin^2 \alpha = \frac{1 - \cos(2\alpha)}{2}$ to rewrite the numerator. Let $\alpha = \pi \cos^4 x$. Then $\sin^2(\pi \cos^4 x) = \frac{1 - \cos(2\pi \cos^4 x)}{2}$. The limit becomes: $\mathop {\lim }\limits_{x \to 0} \frac{\frac{1 - \cos(2\pi \cos^4 x)}{2}}{x^4} = \mathop {\lim }\limits_{x \to 0} \frac{1 - \cos(2\pi \cos^4 x)}{2x^4}$

Step 2: Manipulate the numerator to use the limit form $\mathop {\lim }\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$. As $x \to 0$, $\cos x \to 1$, so $\cos^4 x \to 1$. Therefore, $2\pi \cos^4 x \to 2\pi$. We have a term of the form $1 - \cos(\theta)$ where $\theta \to 2\pi$. This is not directly the standard limit form. However, we can use the property $\cos(2\pi - \phi) = \cos(\phi)$. Let $\phi = 2\pi (1 - \cos^4 x)$. Then $2\pi \cos^4 x = 2\pi (1 - (1 - \cos^4 x)) = 2\pi - 2\pi (1 - \cos^4 x)$. So, $1 - \cos(2\pi \cos^4 x) = 1 - \cos(2\pi - 2\pi (1 - \cos^4 x)) = 1 - \cos(2\pi (1 - \cos^4 x))$. The limit expression is now: $\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos(2\pi (1 - \cos^4 x))}{2x^4}$ To use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$, we need the denominator to be the square of the argument of the cosine. The argument is $2\pi (1 - \cos^4 x)$. So, we multiply and divide by $(2\pi (1 - \cos^4 x))^2$: $\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos(2\pi (1 - \cos^4 x))}{(2\pi (1 - \cos^4 x))^2} \cdot \frac{(2\pi (1 - \cos^4 x))^2}{2x^4}$ This can be broken down into: $\left( \mathop {\lim }\limits_{x \to 0} \frac{1 - \cos(2\pi (1 - \cos^4 x))}{(2\pi (1 - \cos^4 x))^2} \right) \cdot \left( \mathop {\lim }\limits_{x \to 0} \frac{4\pi^2 (1 - \cos^4 x)^2}{2x^4} \right)$ The first limit is of the form $\mathop {\lim }\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$, where $\theta = 2\pi (1 - \cos^4 x)$. As $x \to 0$, $\cos x \to 1$, so $1 - \cos^4 x \to 0$, and thus $\theta \to 0$. So, the first limit is $\frac{1}{2}$.

Step 3: Simplify the remaining expression. The second part of the limit is: $\mathop {\lim }\limits_{x \to 0} \frac{4\pi^2 (1 - \cos^4 x)^2}{2x^4} = 2\pi^2 \mathop {\lim }\limits_{x \to 0} \frac{(1 - \cos^4 x)^2}{x^4}$ We can factor the term $(1 - \cos^4 x)$ as a difference of squares: $1 - \cos^4 x = (1 - \cos^2 x)(1 + \cos^2 x)$. So, $(1 - \cos^4 x)^2 = (1 - \cos^2 x)^2 (1 + \cos^2 x)^2$. Since $1 - \cos^2 x = \sin^2 x$, we have: $(1 - \cos^4 x)^2 = (\sin^2 x)^2 (1 + \cos^2 x)^2 = \sin^4 x (1 + \cos^2 x)^2$ Substitute this back into the limit: $2\pi^2 \mathop {\lim }\limits_{x \to 0} \frac{\sin^4 x (1 + \cos^2 x)^2}{x^4}$ We can rewrite this as: $2\pi^2 \left( \mathop {\lim }\limits_{x \to 0} \frac{\sin^4 x}{x^4} \right) \left( \mathop {\lim }\limits_{x \to 0} (1 + \cos^2 x)^2 \right)$ Using the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$, we have $\mathop {\lim }\limits_{x \to 0} \frac{\sin^4 x}{x^4} = \left(\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}\right)^4 = 1^4 = 1$. For the second part, as $x \to 0$, $\cos x \to 1$, so $\cos^2 x \to 1$. $\mathop {\lim }\limits_{x \to 0} (1 + \cos^2 x)^2 = (1 + 1)^2 = 2^2 = 4$

Step 4: Combine the results from Step 2 and Step 3. The overall limit is the product of the results from the two parts: $\left( \frac{1}{2} \right) \cdot \left( 2\pi^2 \cdot 1 \cdot 4 \right) = \frac{1}{2} \cdot 8\pi^2 = 4\pi^2$

Common Mistakes & Tips

• Incorrect application of standard limits: Ensure the argument of the trigonometric function in the numerator matches the denominator (or its square for cosine terms) and that the limit variable approaches the correct value (usually 0).
• Algebraic errors during simplification: Be meticulous with algebraic manipulations, especially when dealing with powers and factorizations. Double-checking the steps can prevent errors.
• Forgetting the constant factor: When using identities like $\sin^2 \alpha = \frac{1 - \cos(2\alpha)}{2}$, remember to account for the factor of $\frac{1}{2}$ in the overall calculation.

Summary

The problem requires evaluating a limit involving trigonometric functions. We begin by applying the double angle identity for $\sin^2 \alpha$ to transform the expression. Subsequently, we manipulate the terms to align with the standard limit form $\mathop {\lim }\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$. This involves recognizing that $\cos(2\pi - \phi) = \cos(\phi)$ and strategically multiplying and dividing by the appropriate terms. The expression is then simplified by factoring and utilizing the fundamental limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$. Combining the results of these steps leads to the final answer.

The final answer is $\boxed{4\pi^2}$. This corresponds to option (C).