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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.5.69

Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76. 
Try it—you just keep on cycling. Find the limits some other way.
69. lim (x → ∞) (√(9x + 1)) / (√(x + 1))

Guida verificata passo dopo passo
1
Identify the limit expression: \(\lim_{x \to \infty} \frac{\sqrt{9x + 1}}{\sqrt{x + 1}}\).
Since both numerator and denominator involve square roots of expressions that grow without bound as \(x \to \infty\), try to simplify the expression by factoring out \(x\) inside the square roots.
Rewrite the numerator and denominator as \(\sqrt{x \left(9 + \frac{1}{x}\right)}\) and \(\sqrt{x \left(1 + \frac{1}{x}\right)}\) respectively.
Use the property of square roots to separate the factors: \(\frac{\sqrt{x} \sqrt{9 + \frac{1}{x}}}{\sqrt{x} \sqrt{1 + \frac{1}{x}}}\).
Cancel \(\sqrt{x}\) from numerator and denominator, then evaluate the limit of the remaining expression \(\frac{\sqrt{9 + \frac{1}{x}}}{\sqrt{1 + \frac{1}{x}}}\) as \(x \to \infty\) by substituting \(\frac{1}{x} \to 0\).

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