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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 78a

Analyze lim x→∞ f(x) and lim x→−∞ f(x), and then identify any horizontal asymptotes.


f(x)=|1−x^2| / x(x+1)

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The function given is \( f(x) = \frac{|1-x^2|}{x(x+1)} \). We need to analyze the behavior of this function as \( x \to \infty \) and \( x \to -\infty \).
For large values of \( x \), the term \( x^2 \) dominates \( 1 \) in \( |1-x^2| \), so \( |1-x^2| \approx |x^2| = x^2 \). Thus, \( f(x) \approx \frac{x^2}{x(x+1)} = \frac{x^2}{x^2 + x} \).
Simplify \( \frac{x^2}{x^2 + x} \) by dividing the numerator and the denominator by \( x^2 \), resulting in \( \frac{1}{1 + \frac{1}{x}} \). As \( x \to \infty \), \( \frac{1}{x} \to 0 \), so \( \lim_{x \to \infty} f(x) = \frac{1}{1+0} = 1 \).
For \( x \to -\infty \), the simplification is similar: \( f(x) \approx \frac{x^2}{x^2 + x} = \frac{1}{1 + \frac{1}{x}} \). As \( x \to -\infty \), \( \frac{1}{x} \to 0 \), so \( \lim_{x \to -\infty} f(x) = 1 \).
Since both \( \lim_{x \to \infty} f(x) \) and \( \lim_{x \to -\infty} f(x) \) are equal to 1, the horizontal asymptote of the function is \( y = 1 \).

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