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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
2장, 문제 2.6.89

Finding Limits of Differences When x → ±∞


Find the limits in Exercises 84–90. (Hint: Try multiplying and dividing by the conjugate.)


lim x → ∞ (√(x² + 3x) − √(x² − 2x))

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1
Identify the expression whose limit you need to find: \( \lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 - 2x}) \).
To simplify the expression, multiply and divide by the conjugate: \( \frac{(\sqrt{x^2 + 3x} - \sqrt{x^2 - 2x})(\sqrt{x^2 + 3x} + \sqrt{x^2 - 2x})}{\sqrt{x^2 + 3x} + \sqrt{x^2 - 2x}} \).
The numerator becomes a difference of squares: \( (x^2 + 3x) - (x^2 - 2x) = 5x \).
The expression now is \( \frac{5x}{\sqrt{x^2 + 3x} + \sqrt{x^2 - 2x}} \).
Divide both the numerator and the denominator by \( x \) to simplify: \( \frac{5}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 - \frac{2}{x}}} \). Evaluate the limit as \( x \to \infty \).

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주요 개념

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Limits at Infinity

Limits at infinity involve finding the behavior of a function as the variable approaches positive or negative infinity. This concept helps determine the end behavior of functions, which is crucial for understanding asymptotic behavior and horizontal asymptotes. In this context, it helps analyze how the expression behaves as x becomes very large.
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Conjugate multiplication is a technique used to simplify expressions involving square roots. By multiplying the numerator and denominator by the conjugate, you can eliminate the square roots, making it easier to evaluate limits. This method is particularly useful when dealing with differences of square roots, as it transforms the expression into a more manageable form.
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Simplifying expressions involves reducing them to their simplest form, often by factoring, combining like terms, or using algebraic identities. In the context of limits, simplification can reveal the dominant terms that dictate the behavior of the function as x approaches infinity, allowing for easier evaluation of the limit.
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