To evaluate the limit $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$ using the Squeeze Theorem, follow these steps:

Step 1: Identify bounds for $\sin\left(\frac{1}{x}\right)$

The sine function is always bounded between $-1$ and $1$, regardless of its argument: $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$

Step 2: Multiply by $x^2$

Multiply the entire inequality by $x^2$. Since $x^2 \geq 0$, the inequality's direction remains the same: $- x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$

Step 3: Take the limit of the bounds

Now, take the limit of the left and right expressions as $x \to 0$:

\quad \text{and} \quad \lim_{x \to 0} x^2 = 0$$ ### Step 4: Apply the Squeeze Theorem By the **Squeeze Theorem**, since $$-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$$ and both bounds tend to $$0$$ as $$x \to 0$$, the middle expression must also tend to $$0$$. Therefore: $$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$$ Thus, the limit is $$\boxed{0}$$. Would you like further details or have any questions? --- Here are 5 related questions for further exploration: 1. How does the Squeeze Theorem apply to other trigonometric functions? 2. What happens if we replace $$\sin\left(\frac{1}{x}\right)$$ with $$\cos\left(\frac{1}{x}\right)$$? 3. Can we evaluate $$\lim_{x \to 0} x^n \sin\left(\frac{1}{x}\right)$$ for other powers $$n$$? 4. How can you apply the Squeeze Theorem to functions involving exponential limits? 5. How would the result change if we considered the limit as $$x \to \infty$$? ### Tip: The Squeeze Theorem is especially useful when dealing with functions that oscillate but are bounded, like sine and cosine, where direct evaluation isn't possible.