find the limit  y = f(x) =

x

2sin(

1

x2

)

cosx

as x → 0.

We are tasked with finding the limit of the function:

\[
y = f(x) = x \cdot 2\sin\left(\frac{1}{x^2}\right) \cos(x)
\]
as \( x \to 0 \).

### Step-by-step process:

1. **Analyze each part of the function**:
   \[
   f(x) = x \cdot 2\sin\left(\frac{1}{x^2}\right) \cos(x)
   \]
   - \( x \to 0 \)
   - \( 2\sin\left(\frac{1}{x^2}\right) \) oscillates between -2 and 2, but it doesn't have a definite limit as \( x \to 0 \) because the sine function oscillates rapidly as \( \frac{1}{x^2} \to \infty \).
   - \( \cos(x) \to 1 \) as \( x \to 0 \).

2. **Behavior of the product**:
   As \( x \to 0 \), we have:
   - \( x \to 0 \),
   - \( 2\sin\left(\frac{1}{x^2}\right) \) oscillates between -2 and 2,
   - \( \cos(x) \to 1 \).

   So, the term \( 2\sin\left(\frac{1}{x^2}\right) \cos(x) \) oscillates between -2 and 2, but it's multiplied by \( x \), which approaches 0.

3. **Conclusion**:
   Since \( x \cdot 2\sin\left(\frac{1}{x^2}\right) \cos(x) \) is the product of a term that oscillates between -2 and 2 and a term that approaches 0 (i.e., \( x \)), the limit of the product as \( x \to 0 \) is 0.

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

### Final Answer:
\[
\lim_{x \to 0} f(x) = 0
\]

Would you like further details, or do you have any questions about the steps?

Here are 5 related questions to further expand your understanding:

1. How do we handle limits involving oscillating functions like \( \sin\left(\frac{1}{x}\right) \)?
2. What is the Squeeze Theorem and how can it apply here?
3. What happens if we modify the function by replacing \( \sin\left(\frac{1}{x^2}\right) \) with \( \cos\left(\frac{1}{x^2}\right) \)?
4. How would the limit change if we removed the cosine term?
5. How do limits behave when oscillations are involved in more complex functions?

**Tip:** When dealing with oscillating functions multiplied by terms approaching zero, the Squeeze Theorem is a powerful tool for proving that the overall limit is zero.

find the limit y = f(x) = x 2sin(1 / x^2) cosx as x → 0.

Discover how to find the limit of the function f(x) = x * 2sin(1/x^2) * cos(x) as x approaches 0. This problem involves the Squeeze Theorem and the behavior of oscillating functions. Ideal for college-level calculus students.

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