Math Problem Statement
Solution
The expression you're dealing with is:
Step-by-Step Solution:
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Examine the behavior of the numerator as :
- as , by using the small angle approximation: for small .
- as , using the same approximation for .
So, the numerator becomes: However, this approximation only gives us a first-order idea. We need to consider higher-order terms for more accuracy.
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Taylor Expansions:
- The Taylor series of around is:
- The Taylor series of around is:
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Simplify the numerator: Using these expansions, we get: Simplifying this:
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Substitute into the limit expression:
Now the expression for the limit becomes:
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Simplify the limit:
Dividing term-by-term by , we get:
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Evaluate the limit:
As , all terms involving higher powers of vanish, and we're left with:
Would you like further clarifications on any step?
Related Questions:
- How does the Taylor expansion help in evaluating limits?
- What are common small-angle approximations and when should they be used?
- What would happen if we considered only first-order approximations in this problem?
- Could L'Hopital's rule be applied to solve this limit instead?
- How does this problem relate to the general behavior of trigonometric functions near zero?
Tip:
When dealing with limits involving trigonometric functions, Taylor expansions are a powerful tool to capture the behavior near the point of interest, especially for higher-order terms.
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Math Problem Analysis
Mathematical Concepts
Limits
Taylor Series Expansion
Trigonometric Functions
Formulas
sin(x) ≈ x for small x
Taylor expansion of sin(x^2) = x^2 - x^6/6 + O(x^8)
Taylor expansion of sin^2(x) = x^2 - x^4/3 + O(x^6)
Theorems
Small Angle Approximation
Taylor Series
Suitable Grade Level
University Level Calculus