Math Problem Statement

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

Solution

The inequality 1<sin(x)<1-1 < \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 xx, given that the sine function is always constrained between 1-1 and 11.

Squeeze Theorem

The Squeeze Theorem states that if:

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

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

In your example of 1<sin(x)<1-1 < \sin(x) < 1, the functions 1-1 and 11 are constants, and they serve as the bounds for sin(x)\sin(x), but no limit comparison is being made to determine the behavior at a particular point. Instead, sin(x)\sin(x) simply oscillates between 1-1 and 11, 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:
    • limxa1=1\lim_{x \to a} -1 = -1,
    • limxa1=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)<1-1 < \sin(x) < 1 is just the natural range of the sine function, a fact that holds for all xx, and not a limit statement.

Summary

The inequality 1<sin(x)<1-1 < \sin(x) < 1 doesn't conform to the Squeeze Theorem because the upper and lower bounds 1-1 and 11 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)\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 xx \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.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Limits
Squeeze Theorem

Formulas

-1 < sin(x) < 1
Squeeze Theorem conditions

Theorems

Squeeze Theorem

Suitable Grade Level

Grades 11-12 (Advanced Calculus)