Key Concepts and Formulas

• Trigonometric Identities:
  • Cosine subtraction formula: $\cos A - \cos B = 2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)$
• Taylor Series Expansions (around $x=0$):
  • $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
  • $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
• Standard Limits:
  • $\lim_{x \to 0} \frac{\sin x}{x} = 1$
  • $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$

Step-by-Step Solution

Step 1: Apply the cosine subtraction formula to the numerator. We are given the limit: $\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}}$ We can use the identity $\cos A - \cos B = 2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)$ with $A = \sin x$ and $B = x$. This transforms the numerator as follows: $\cos(\sin x) - \cos x = 2 \sin \left(\frac{\sin x + x}{2}\right) \sin \left(\frac{x - \sin x}{2}\right)$ So the limit becomes: $\mathop {\lim }\limits_{x \to 0} {{2\sin \left(\frac{x + \sin x}{2}\right) \sin \left(\frac{x - \sin x}{2}\right)} \over {{x^4}}}$

Step 2: Rearrange the terms to utilize standard limits. To make use of the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$, we need to divide and multiply by appropriate terms. We can rewrite the expression as: $\mathop {\lim }\limits_{x \to 0} 2 \cdot \left(\frac{\sin \left(\frac{x + \sin x}{2}\right)}{\frac{x + \sin x}{2}}\right) \cdot \left(\frac{\sin \left(\frac{x - \sin x}{2}\right)}{\frac{x - \sin x}{2}}\right) \cdot \left(\frac{\frac{x + \sin x}{2}}{1}\right) \cdot \left(\frac{\frac{x - \sin x}{2}}{1}\right) \cdot \frac{1}{x^4}$ As $x \to 0$, we have $\sin x \to 0$, so $\frac{x + \sin x}{2} \to 0$ and $\frac{x - \sin x}{2} \to 0$. Thus, $\lim_{x \to 0} \frac{\sin \left(\frac{x + \sin x}{2}\right)}{\frac{x + \sin x}{2}} = 1$ and $\lim_{x \to 0} \frac{\sin \left(\frac{x - \sin x}{2}\right)}{\frac{x - \sin x}{2}} = 1$. The limit simplifies to: $\mathop {\lim }\limits_{x \to 0} 2 \cdot 1 \cdot 1 \cdot \left(\frac{x + \sin x}{2}\right) \cdot \left(\frac{x - \sin x}{2}\right) \cdot \frac{1}{x^4}$ $= \mathop {\lim }\limits_{x \to 0} 2 \cdot \frac{(x + \sin x)(x - \sin x)}{4x^4}$ $= \mathop {\lim }\limits_{x \to 0} \frac{1}{2} \cdot \frac{x^2 - \sin^2 x}{x^4}$

Step 3: Use Taylor series expansions for $\sin x$. As $x \to 0$, we can use the Taylor series expansion for $\sin x$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$ Then, $\sin^2 x = \left(x - \frac{x^3}{6} + \dots\right)^2 = x^2 - 2x \left(\frac{x^3}{6}\right) + \left(\frac{x^3}{6}\right)^2 + \dots$ $\sin^2 x = x^2 - \frac{x^4}{3} + O(x^6)$ Substitute this into the expression from Step 2: $\mathop {\lim }\limits_{x \to 0} \frac{1}{2} \cdot \frac{x^2 - \left(x^2 - \frac{x^4}{3} + O(x^6)\right)}{x^4}$ $= \mathop {\lim }\limits_{x \to 0} \frac{1}{2} \cdot \frac{x^2 - x^2 + \frac{x^4}{3} - O(x^6)}{x^4}$ $= \mathop {\lim }\limits_{x \to 0} \frac{1}{2} \cdot \frac{\frac{x^4}{3} - O(x^6)}{x^4}$

Step 4: Simplify and evaluate the limit. Divide the numerator and denominator by $x^4$: $= \mathop {\lim }\limits_{x \to 0} \frac{1}{2} \cdot \left(\frac{1}{3} - O(x^2)\right)$ As $x \to 0$, the term $O(x^2)$ goes to zero. $= \frac{1}{2} \cdot \frac{1}{3}$ $= \frac{1}{6}$

Alternative Step 3 (Using Taylor Series for $\cos x$ and $\cos(\sin x)$): We can also solve this using Taylor series expansions for $\cos x$ and $\cos(\sin x)$ directly. The Taylor expansion for $\cos u$ around $u=0$ is $\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots$. For $\cos x$: $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$ For $\cos(\sin x)$, we substitute the Taylor series of $\sin x$ into the Taylor series of $\cos u$. $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$ Let $u = \sin x$. As $x \to 0$, $u \to 0$. $\cos(\sin x) = 1 - \frac{(\sin x)^2}{2} + \frac{(\sin x)^4}{24} - \dots$ Substitute the expansion of $\sin x$: $\cos(\sin x) = 1 - \frac{1}{2}\left(x - \frac{x^3}{6} + \dots\right)^2 + \frac{1}{24}\left(x - \frac{x^3}{6} + \dots\right)^4 - \dots$ $\cos(\sin x) = 1 - \frac{1}{2}\left(x^2 - 2x\left(\frac{x^3}{6}\right) + \dots\right) + \frac{1}{24}\left(x^4 + \dots\right) - \dots$ $\cos(\sin x) = 1 - \frac{1}{2}\left(x^2 - \frac{x^4}{3} + \dots\right) + \frac{x^4}{24} - \dots$ $\cos(\sin x) = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} + \dots$ $\cos(\sin x) = 1 - \frac{x^2}{2} + \left(\frac{1}{6} + \frac{1}{24}\right)x^4 + \dots$ $\cos(\sin x) = 1 - \frac{x^2}{2} + \frac{5}{24}x^4 + \dots$ Now, consider the numerator: $\cos(\sin x) - \cos x$. $\cos(\sin x) - \cos x = \left(1 - \frac{x^2}{2} + \frac{5}{24}x^4 + \dots\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots\right)$ $= \left(1 - 1\right) + \left(-\frac{x^2}{2} - (-\frac{x^2}{2})\right) + \left(\frac{5}{24}x^4 - \frac{1}{24}x^4\right) + \dots$ $= 0 + 0 + \frac{4}{24}x^4 + \dots$ $= \frac{1}{6}x^4 + O(x^6)$ Now, substitute this back into the limit: $\mathop {\lim }\limits_{x \to 0} \frac{\frac{1}{6}x^4 + O(x^6)}{x^4} = \mathop {\lim }\limits_{x \to 0} \left(\frac{1}{6} + O(x^2)\right)$ $= \frac{1}{6}$

Common Mistakes & Tips

• Incorrect Taylor Series: Ensure you are using the correct Taylor series expansions for $\sin x$ and $\cos x$, and that you are expanding them to a sufficient order to cancel out terms in the numerator. For a denominator of $x^4$, you typically need terms up to $x^4$ in the numerator's expansion.
• Algebraic Errors: Be very careful with algebraic manipulations, especially when squaring terms or combining fractions. A small error can lead to a completely wrong answer.
• L'Hôpital's Rule: While L'Hôpital's rule can be applied here, it would require differentiating the numerator and denominator four times, which is tedious and prone to errors. Using Taylor series or trigonometric identities is generally more efficient for this type of limit.
• Approximation: When using Taylor series, remember that terms of higher order than needed can be ignored (represented by $O(x^n)$) as they tend to zero faster than the dominant terms as $x \to 0$.

Summary

The problem requires evaluating a limit of the form $\frac{0}{0}$. We can solve this by applying the cosine subtraction formula and then using Taylor series expansions for $\sin x$ or by directly expanding $\cos(\sin x)$ and $\cos x$ around $x=0$. Both methods involve carefully expanding the trigonometric functions to a sufficient order to cancel out the lower-order terms in the numerator. After simplification, the limit evaluates to $\frac{1}{6}$.

The final answer is $\boxed{{1 \over 6}}$.