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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 2.75a

Analyze lim x→∞ f(x) and lim x→−∞ f(x), and then identify any horizontal asymptotes.
f(x) = (x2 − 9)/(x(x−3))

Guida verificata passo dopo passo
1
Start by simplifying the function \( f(x) = \frac{x^2 - 9}{x(x-3)} \). Notice that the numerator \( x^2 - 9 \) can be factored as \((x-3)(x+3)\). So, the function becomes \( f(x) = \frac{(x-3)(x+3)}{x(x-3)} \).
Cancel the common factor \( (x-3) \) from the numerator and the denominator, but remember that \( x \neq 3 \) to avoid division by zero. The simplified function is \( f(x) = \frac{x+3}{x} \).
To find the limit as \( x \to \infty \), consider \( f(x) = \frac{x+3}{x} = 1 + \frac{3}{x} \). As \( x \to \infty \), \( \frac{3}{x} \to 0 \), so \( \lim_{x \to \infty} f(x) = 1 \).
Similarly, for \( x \to -\infty \), \( f(x) = 1 + \frac{3}{x} \). As \( x \to -\infty \), \( \frac{3}{x} \to 0 \), so \( \lim_{x \to -\infty} f(x) = 1 \).
Since both \( \lim_{x \to \infty} f(x) \) and \( \lim_{x \to -\infty} f(x) \) equal 1, the horizontal asymptote of the function is \( y = 1 \).

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