To determine which step helps to show $\lim_{x \to 0} x^2 \sin\left(\frac{3}{x}\right) = 0$ using the squeeze theorem, let's break down the problem.

Step 1: Understanding the Behavior of $\sin\left(\frac{3}{x}\right)$

The function $\sin\left(\frac{3}{x}\right)$ oscillates between $-1$ and $1$ for all values of $x$. This gives us the inequality: $-1 \leq \sin\left(\frac{3}{x}\right) \leq 1$

Step 2: Multiplying the inequality by $x^2$

Since $x^2 \geq 0$ for all $x$, multiplying all parts of the inequality by $x^2$ gives: $- x^2 \leq x^2 \sin\left(\frac{3}{x}\right) \leq x^2$

Step 3: Applying the Squeeze Theorem

Now we take the limit of all parts of this inequality as $x \to 0$: $\lim_{x \to 0} -x^2 = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0$

Since both the left-hand side and the right-hand side limits are 0, by the squeeze theorem, we conclude: $\lim_{x \to 0} x^2 \sin\left(\frac{3}{x}\right) = 0$

Correct Step

The correct step is recognizing that $-1 \leq \sin\left(\frac{3}{x}\right) \leq 1$ and multiplying it by $x^2$ to squeeze the original function $x^2 \sin\left(\frac{3}{x}\right)$.

Would you like more details or have any questions?

Here are some related questions to expand on this concept:

1. How does the oscillating nature of $\sin(x)$ affect limit problems like this?
2. What is the formal definition of the squeeze theorem?
3. Why can't we directly compute the limit of $\sin\left(\frac{3}{x}\right)$ as $x \to 0$?
4. Are there other trigonometric functions for which the squeeze theorem is useful in limits?
5. How does the behavior of $x^2$ help in making the limit go to 0?

Tip: The squeeze theorem is especially useful for limits involving oscillating functions like sine or cosine.