how does -1<sinx<1 conform to the squeeze theorem if lim -1 does not = lim 1

The inequality \(-1 < \sin(x) < 1\) does not directly invoke the Squeeze Theorem because it's not a statement about limits in the conventional sense. Instead, it's simply a bound that holds for all real numbers \(x\), given that the sine function is always constrained between \(-1\) and \(1\).

### Squeeze Theorem
The **Squeeze Theorem** states that if:

1. \( g(x) \leq f(x) \leq h(x) \) for all \(x\) near a point \(a\) (except possibly at \(a\)),
2. \( \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \),

then, \( \lim_{x \to a} f(x) = L \).

In your example of \(-1 < \sin(x) < 1\), the functions \(-1\) and \(1\) are **constants**, and they serve as the bounds for \(\sin(x)\), but no limit comparison is being made to determine the behavior at a particular point. Instead, \(\sin(x)\) simply oscillates between \(-1\) and \(1\), never exceeding these bounds.

### Why Limit Comparison Doesn’t Work Here:
- For the Squeeze Theorem to work, the upper and lower bounds must converge to the same value (limit) at a specific point. However, in this case:
  - \( \lim_{x \to a} -1 = -1 \),
  - \( \lim_{x \to a} 1 = 1 \).

Since these limits are not equal, the Squeeze Theorem is not applicable in this context.

Instead, \(-1 < \sin(x) < 1\) is just the natural range of the sine function, a fact that holds for all \(x\), and not a limit statement.

### Summary
The inequality \(-1 < \sin(x) < 1\) doesn't conform to the Squeeze Theorem because the upper and lower bounds \(-1\) and \(1\) are constant, and their limits do not match. The Squeeze Theorem applies when the bounds converge to the same value at a particular point.

Would you like further clarification or details on this? Here are five related questions to deepen understanding:

1. What is an example of the Squeeze Theorem applied to a limit problem?
2. Can \(\sin(x)\) be used in a Squeeze Theorem scenario for specific limits?
3. How does the range of sine and cosine functions compare to other periodic functions?
4. What happens to the sine function as \(x \to \infty\)?
5. What are common applications of the Squeeze Theorem in calculus?

**Tip:** The Squeeze Theorem is often used to evaluate limits involving trigonometric functions, especially when the function oscillates or has complex behavior near a certain point.

Explore how the inequality -1 < sin(x) < 1 relates to the Squeeze Theorem and why it does not conform due to limit differences. This problem covers key concepts in trigonometry, limits, and the Squeeze Theorem. Suitable for advanced calculus students in Grades 11-12.

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