To investigate the continuity of the function $y(x) = x^2 \sin\left(\frac{1}{x}\right)$, we need to analyze its behavior at different points, particularly around $x = 0$, since this is where potential discontinuity or undefined behavior might occur.

Step 1: Determine the Domain

The function $y(x) = x^2 \sin\left(\frac{1}{x}\right)$ is defined for all $x \neq 0$. At $x = 0$, $\frac{1}{x}$ is undefined, so we have a problem of continuity at $x = 0$.

Step 2: Continuity at $x \neq 0$

For any $x \neq 0$, the function $\sin\left(\frac{1}{x}\right)$ is defined, and since $x^2$ is also defined for all real numbers, the function $y(x) = x^2 \sin\left(\frac{1}{x}\right)$ is continuous for all $x \neq 0$.

Step 3: Investigate Continuity at $x = 0$

To determine whether the function is continuous at $x = 0$, we need to check if $\lim_{x \to 0} y(x)$ exists and if it equals $y(0)$. Since $y(x)$ is not defined at $x = 0$, we consider the limit:

$\lim_{x \to 0} y(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$

Step 4: Evaluate the Limit as $x \to 0$

Since $\sin\left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$, we can bound $y(x)$ as follows:

$- x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$

As $x \to 0$, both $x^2$ and $-x^2$ approach 0. By the Squeeze Theorem, this implies:

$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$

Step 5: Define $y(0)$ to Ensure Continuity

If we define $y(0) = 0$, then:

$\lim_{x \to 0} y(x) = y(0) = 0$

This makes the function continuous at $x = 0$.

Conclusion

The function $y(x) = x^2 \sin\left(\frac{1}{x}\right)$ is continuous for all $x$ if we define $y(0) = 0$.

Would you like further details on any part of this explanation, or do you have any additional questions?

Here are some related questions that might help deepen your understanding:

1. How does the Squeeze Theorem work, and why is it applicable here?
2. What happens if $x$ approaches zero for other types of functions involving $\sin\left(\frac{1}{x}\right)$?
3. Why do oscillatory functions like $\sin\left(\frac{1}{x}\right)$ cause complications in continuity analysis?
4. Can you find the derivative of $y(x) = x^2 \sin\left(\frac{1}{x}\right)$ and discuss its continuity?
5. What are other examples of functions with discontinuities that can be made continuous by defining them at certain points?

Tip: When dealing with oscillatory functions like $\sin\left(\frac{1}{x}\right)$, the Squeeze Theorem can often help in determining limits as $x \to 0$ by bounding the oscillations.