We are asked to evaluate the following limit using the Squeeze Theorem:

$\lim_{x \to 0} \sqrt{x^3 + x^2} \sin\left(\frac{\pi}{x}\right).$

Step 1: Understand the behavior of the sine function

The term $\sin\left(\frac{\pi}{x}\right)$ oscillates between $-1$ and $1$ for all values of $x \neq 0$:

$-1 \leq \sin\left(\frac{\pi}{x}\right) \leq 1.$

Multiplying by $\sqrt{x^3 + x^2}$, we have:

$-\sqrt{x^3 + x^2} \leq \sqrt{x^3 + x^2} \sin\left(\frac{\pi}{x}\right) \leq \sqrt{x^3 + x^2}.$

Step 2: Apply limits to the bounds

Now, we calculate the limit of the bounding functions as $x \to 0$.

Limit of $\sqrt{x^3 + x^2}$ as $x \to 0$:

$\sqrt{x^3 + x^2} = \sqrt{x^2(x + 1)} = |x|\sqrt{x + 1}.$

As $x \to 0$:

$|x|\sqrt{x + 1} \to 0.$

Thus:

$\lim_{x \to 0} \sqrt{x^3 + x^2} = 0.$

Step 3: Apply the Squeeze Theorem

Since:

$-\sqrt{x^3 + x^2} \leq \sqrt{x^3 + x^2} \sin\left(\frac{\pi}{x}\right) \leq \sqrt{x^3 + x^2},$

and both the lower and upper bounds tend to 0 as $x \to 0$, by the Squeeze Theorem, we conclude:

$\lim_{x \to 0} \sqrt{x^3 + x^2} \sin\left(\frac{\pi}{x}\right) = 0.$

Final Answer:

$\boxed{0}.$

Would you like me to explain the Squeeze Theorem further or go over related concepts?

Related Questions for Practice:

1. What is the Squeeze Theorem, and when is it typically used in calculus?
2. How does the behavior of the $\sin\left(\frac{\pi}{x}\right)$ function affect the limit process?
3. Why is the limit of $\sqrt{x^3 + x^2}$ important in determining the final result?
4. How would the result change if the expression inside the square root were different (e.g., $x^2 + x$ instead)?
5. How do you handle limits involving oscillating functions like $\sin$ or $\cos$?

Tip:

When applying the Squeeze Theorem, always ensure that the function is bounded between two other functions whose limits are known or easier to compute.