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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
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7장, 문제 7.PE.95

Use l’Hôpital’s Rule to find the limits in Exercises 85–108.
95. lim(x→∞) (√(x² + x + 1) - √(x² - x))

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Identify the limit expression: \(\lim_{x \to \infty} \left( \sqrt{x^{2} + x + 1} - \sqrt{x^{2} - x} \right)\).
Recognize that as \(x \to \infty\), both \(\sqrt{x^{2} + x + 1}\) and \(\sqrt{x^{2} - x}\) behave like \(x\), so the expression is of the indeterminate form \(\infty - \infty\).
To apply l’Hôpital’s Rule, first rewrite the expression to a quotient form that yields an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Multiply and divide by the conjugate to rationalize the expression: multiply numerator and denominator by \(\sqrt{x^{2} + x + 1} + \sqrt{x^{2} - x}\).
Simplify the numerator using the difference of squares formula: \((a - b)(a + b) = a^{2} - b^{2}\), which gives \(\left( x^{2} + x + 1 \right) - \left( x^{2} - x \right) = 2x + 1\).
Rewrite the limit as \(\lim_{x \to \infty} \frac{2x + 1}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} - x}}\) and then analyze this new expression to find the limit, possibly applying l’Hôpital’s Rule if the form is still indeterminate.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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