f(x)=x^2 sin(1/x^2) divide by cos x and limit x approaches 0

We are tasked with evaluating the following limit:

\[
\lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x^2}\right)}{\cos(x)}
\]

### Step-by-Step Approach:

1. **Behavior of \(\cos(x)\) as \(x \to 0\)**:
   - We know that \(\cos(x) \to 1\) as \(x \to 0\).

2. **Behavior of \(x^2 \sin\left(\frac{1}{x^2}\right)\) as \(x \to 0\)**:
   - \(\sin\left(\frac{1}{x^2}\right)\) oscillates between -1 and 1 for all values of \(x\), because \(\sin\) is a bounded function.
   - Therefore, \(x^2 \sin\left(\frac{1}{x^2}\right)\) is also bounded, but its magnitude is always less than or equal to \(x^2\), i.e.,
     \[
     -x^2 \leq x^2 \sin\left(\frac{1}{x^2}\right) \leq x^2
     \]
   - As \(x \to 0\), \(x^2 \to 0\). By the **squeeze theorem**, we can conclude that
     \[
     \lim_{x \to 0} x^2 \sin\left(\frac{1}{x^2}\right) = 0
     \]

3. **Limit of the entire expression**:
   - Using the fact that \(\cos(x) \to 1\) as \(x \to 0\) and from the squeeze theorem, we conclude that the numerator \(x^2 \sin\left(\frac{1}{x^2}\right)\) approaches 0. Therefore, the whole expression approaches:
     \[
     \lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x^2}\right)}{\cos(x)} = \frac{0}{1} = 0
     \]

### Final Answer:
\[
\lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x^2}\right)}{\cos(x)} = 0
\]

Would you like any further details or clarifications?

Here are 5 related questions to deepen your understanding:
1. How does the squeeze theorem work, and when is it applicable?
2. Why does \(\sin\left(\frac{1}{x^2}\right)\) oscillate as \(x\) approaches 0?
3. Can you evaluate \(\lim_{x \to 0} \frac{\sin(x)}{x}\)?
4. How do bounded functions help in determining limits?
5. What is the difference between an oscillating and non-oscillating function as \(x\) approaches 0?

**Tip:** When dealing with limits of oscillating functions, the squeeze theorem is often a powerful tool to simplify the problem.

f(x) = x^2 sin(1/x^2) divided by cos x and limit as x approaches 0

Explore the limit problem involving oscillating functions and the squeeze theorem: lim_{x → 0} (x^2 sin(1/x^2)) / cos(x). Learn how the behavior of trigonometric functions and bounded behavior lead to the final solution using calculus techniques. Ideal for Calculus I students.

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