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The expression you're dealing with is:

\[
\lim_{x \to 0} \frac{\sin(x^2) - \sin^2(x)}{x^4}
\]

### Step-by-Step Solution:

1. **Examine the behavior of the numerator as \(x \to 0\):**

   - \(\sin(x^2) \approx x^2\) as \(x \to 0\), by using the small angle approximation: \(\sin y \approx y\) for small \(y\).
   - \(\sin^2(x) \approx x^2\) as \(x \to 0\), using the same approximation for \(\sin(x)\).
   
   So, the numerator becomes:
   \[
   \sin(x^2) - \sin^2(x) \approx x^2 - x^2 = 0.
   \]
   However, this approximation only gives us a first-order idea. We need to consider higher-order terms for more accuracy.

2. **Taylor Expansions:**

   - The Taylor series of \(\sin(x^2)\) around \(x = 0\) is:
     \[
     \sin(x^2) = x^2 - \frac{(x^2)^3}{3!} + O(x^6) = x^2 - \frac{x^6}{6} + O(x^8)
     \]
   - The Taylor series of \(\sin^2(x)\) around \(x = 0\) is:
     \[
     \sin(x) = x - \frac{x^3}{3!} + O(x^5) \implies \sin^2(x) = x^2 - \frac{x^4}{3} + O(x^6)
     \]
     
3. **Simplify the numerator:**
   Using these expansions, we get:
   \[
   \sin(x^2) - \sin^2(x) = \left(x^2 - \frac{x^6}{6}\right) - \left(x^2 - \frac{x^4}{3}\right)
   \]
   Simplifying this:
   \[
   \sin(x^2) - \sin^2(x) = -\frac{x^4}{3} - \frac{x^6}{6} + O(x^8)
   \]

4. **Substitute into the limit expression:**

   Now the expression for the limit becomes:
   \[
   \lim_{x \to 0} \frac{-\frac{x^4}{3} - \frac{x^6}{6} + O(x^8)}{x^4}
   \]

5. **Simplify the limit:**

   Dividing term-by-term by \(x^4\), we get:
   \[
   \lim_{x \to 0} \left(-\frac{1}{3} - \frac{x^2}{6} + O(x^4)\right)
   \]

6. **Evaluate the limit:**

   As \(x \to 0\), all terms involving higher powers of \(x\) vanish, and we're left with:
   \[
   \boxed{-\frac{1}{3}}
   \]

---

Would you like further clarifications on any step?

### Related Questions:
1. How does the Taylor expansion help in evaluating limits?
2. What are common small-angle approximations and when should they be used?
3. What would happen if we considered only first-order approximations in this problem?
4. Could L'Hopital's rule be applied to solve this limit instead?
5. How does this problem relate to the general behavior of trigonometric functions near zero?

### Tip:
When dealing with limits involving trigonometric functions, Taylor expansions are a powerful tool to capture the behavior near the point of interest, especially for higher-order terms.

lim(x → 0) [(sin(x^2) - sin^2(x)) / x^4]

Explore the step-by-step solution to the limit problem lim(x → 0) [(sin(x^2) - sin^2(x)) / x^4] using Taylor series expansions and small angle approximations. Understand the behavior of trigonometric functions near zero. Suitable for university-level calculus students.

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