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.