To calculate the limit:

$\lim_{x \to 0} \frac{\cos(x^2) - 1}{(\sin x)^4}$

Step 1: Recall trigonometric approximations

• $\cos(y) \approx 1 - \frac{y^2}{2}$ as $y \to 0$,
• $\sin(x) \approx x$ as $x \to 0$.

Step 2: Approximate the numerator

Using $\cos(x^2) \approx 1 - \frac{(x^2)^2}{2} = 1 - \frac{x^4}{2}$, the numerator becomes: $\cos(x^2) - 1 \approx \left(1 - \frac{x^4}{2}\right) - 1 = -\frac{x^4}{2}.$

Step 3: Approximate the denominator

For $(\sin x)^4 \approx (x)^4$, since $\sin(x) \approx x$, the denominator becomes: $(\sin x)^4 \approx x^4.$

Step 4: Simplify the limit

The expression simplifies to: $\lim_{x \to 0} \frac{-\frac{x^4}{2}}{x^4}.$

Cancel $x^4$ in the numerator and denominator: $\lim_{x \to 0} \frac{-\frac{1}{2}}{1} = -\frac{1}{2}.$

Final Answer:

$\lim_{x \to 0} \frac{\cos(x^2) - 1}{(\sin x)^4} = -\frac{1}{2}.$

Let me know if you'd like further clarifications or additional steps!

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Related Questions:

1. How do small-angle approximations work for trigonometric functions?
2. What are common strategies for handling indeterminate forms in limits?
3. How is L'Hôpital's rule applied to similar trigonometric limits?
4. Can Taylor series expansions provide alternate solutions to this problem?
5. What are the differences between one-sided and two-sided limits?

Tip:

When working with trigonometric functions, always check if small-angle approximations can simplify the problem. They save time and avoid unnecessary complications.