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The image you've uploaded presents a problem about evaluating a limit using the Squeeze Theorem, particularly involving the sine function. Here's a breakdown of the steps and solutions:

1. **Squeeze Theorem Explanation:**
   - We are given that \( f(x) \leq g(x) \leq h(x) \) for \( x \) in some interval around \( a \), except possibly at \( a \). 
   - If \( \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L \), then \( \lim_{x \to a} g(x) = L \).

2. **The specific problem:**
   - You are asked to use the Squeeze Theorem to evaluate \( \lim_{x \to 0} x^2 \sin \left( \frac{5}{x^2} \right) \).
   - The goal is to find bounds for \( x^2 \sin \left( \frac{5}{x^2} \right) \) and then apply the Squeeze Theorem.

### Step-by-Step Solution:

#### Step 1: **Recognizing the Bound on Sine Function**
   - Recall that for any real number \( y \), the sine function satisfies:
     \[
     -1 \leq \sin(y) \leq 1
     \]
     Therefore, for \( y = \frac{5}{x^2} \), we have:
     \[
     -1 \leq \sin\left( \frac{5}{x^2} \right) \leq 1
     \]

#### Step 2: **Multiplying by \( x^2 \)**
   - Now, multiply the inequality by \( x^2 \), which is nonnegative for all \( x \). This gives:
     \[
     -x^2 \leq x^2 \sin\left( \frac{5}{x^2} \right) \leq x^2
     \]

#### Step 3: **Applying the Squeeze Theorem**
   - From the inequality, we now know that:
     \[
     -x^2 \leq x^2 \sin\left( \frac{5}{x^2} \right) \leq x^2
     \]
   - As \( x \to 0 \), both \( -x^2 \) and \( x^2 \) tend to 0. Hence, by the Squeeze Theorem:
     \[
     \lim_{x \to 0} x^2 \sin\left( \frac{5}{x^2} \right) = 0
     \]

### Conclusion:
   - The limit is \( 0 \).

Let me know if you'd like further clarifications on this or help with another part of the problem!

### Relative Questions:
1. How do you identify functions that are appropriate for the Squeeze Theorem?
2. What are common pitfalls when applying the Squeeze Theorem to trigonometric functions?
3. Can the Squeeze Theorem be used for limits that approach infinity? How?
4. What happens if the outer bounds of the Squeeze Theorem don't tend to the same limit?
5. How does the behavior of \( \sin \left( \frac{1}{x} \right) \) change for small values of \( x \)?

### Tip:
Always check that the functions on both sides of your inequality have the same limit when using the Squeeze Theorem; otherwise, it cannot be applied.

Use the Squeeze Theorem to evaluate the limit of x^2 * sin(5/x^2) as x approaches 0.

Learn how to use the Squeeze Theorem to evaluate the limit of x^2 * sin(5/x^2) as x approaches 0. This problem involves bounding a trigonometric function and applying a core theorem in calculus. Ideal for high school and college calculus students.

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