Math Problem Statement
limx→0+ (1−cosx)^sinx
Solution
To evaluate , let us break it into manageable parts.
Step-by-step solution:
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Rewrite the expression:
Taking the natural logarithm on both sides:
The limit now becomes:
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Analyze : As , approaches . Using the approximation for small : Thus:
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Substitute back: Substituting this into , we have:
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Analyze : As , . Substituting:
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Evaluate each term:
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For : Using L’Hôpital’s Rule:
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For :
Thus, .
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Exponentiate to get the result:
Final Answer:
Would you like a deeper explanation of any step?
Here are five related questions to expand your understanding:
- How is L’Hôpital’s Rule applied in limits involving indeterminate forms?
- Why is a good approximation for small ?
- Can you explain why dominates as ?
- What happens if the base in this limit were different, e.g., ?
- How does Taylor expansion help in solving limits involving trigonometric functions?
Tip: For exponential and logarithmic forms in limits, consider logarithmic properties to simplify calculations.
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Math Problem Analysis
Mathematical Concepts
Limits
Exponential and Logarithmic Functions
Trigonometric Approximations
Indeterminate Forms
Formulas
ln(a^b) = b * ln(a)
1 - cos(x) ≈ x^2 / 2 for small x
L'Hôpital's Rule for indeterminate forms of type 0/0
Theorems
L'Hôpital's Rule
Taylor Expansion for Trigonometric Functions
Natural Logarithm Properties
Suitable Grade Level
Undergraduate Calculus or Advanced High School Calculus