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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 2.6.7

Finding Limits


In Exercises 3–8, find the limit of each function (a) as x → ∞ and (b) as x → −∞. (You may wish to visualize your answer with a graphing calculator or computer.)


h(x) = (−5 + (7/x))/(3 – (1/x²))

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1
Identify the dominant terms in the numerator and the denominator as x approaches infinity. For h(x) = \( \frac{-5 + \frac{7}{x}}{3 - \frac{1}{x^2}} \), the dominant terms are -5 in the numerator and 3 in the denominator.
As x approaches infinity, the terms \( \frac{7}{x} \) and \( \frac{1}{x^2} \) approach zero. Therefore, the expression simplifies to \( \frac{-5}{3} \).
Thus, the limit of h(x) as x approaches infinity is \( \frac{-5}{3} \).
Now, consider the limit as x approaches negative infinity. Again, the terms \( \frac{7}{x} \) and \( \frac{1}{x^2} \) approach zero, simplifying the expression to \( \frac{-5}{3} \).
Therefore, the limit of h(x) as x approaches negative infinity is also \( \frac{-5}{3} \).

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