Math Problem Statement
question c
Solution
To solve these limits using power series, we can expand the functions in the numerator and denominator using their Taylor series at and then evaluate the limit based on the leading terms.
Let's go through each part one by one.
Part (a)
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Expand in a Taylor series around :
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Substitute this expansion into the expression: Simplify by canceling terms:
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Rewrite the limit:
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Divide each term by :
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Take the limit as :
So, the answer for part (a) is:
Part (b)
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Expand as before: So,
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Rewrite the limit:
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Factor out from the denominator:
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Take the limit as :
So, the answer for part (b) is:
Part (c)
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Expand using its Taylor series at :
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Expand using the binomial series around : Therefore,
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Rewrite the limit:
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Factor from both the numerator and denominator:
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Take the limit as :
So, the answer for part (c) is:
Would you like further clarification on any of these parts?
Follow-Up Questions:
- How can power series approximations be applied in calculus beyond limit calculations?
- What are the conditions required for a power series to converge?
- How does the Taylor series differ from the Maclaurin series?
- Can power series be used to approximate values for non-polynomial functions?
- Why is factoring useful when simplifying expressions involving power series?
Tip:
Remember, in limit problems, look for ways to factor or cancel terms to simplify calculations, especially with power series.
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Math Problem Analysis
Mathematical Concepts
Limits
Power Series
Taylor Series
Formulas
ln(1 + x) ≈ x - x^2 / 2 + x^3 / 3 - ...
sqrt(4 + x) ≈ 2 + x/4 - x^2/64 + ...
Theorems
Taylor Series Expansion
Suitable Grade Level
Grades 11-12, College Calculus
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