Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
To determine which functions have horizontal asymptotes, we'll analyze the limits of the given functions as [tex]\( x \)[/tex] approaches infinity.
1. Function [tex]\( f(x)=\frac{x^3-2 x+3}{x^2-5} \)[/tex]
To find the horizontal asymptote, consider the degrees of the numerator and the denominator:
[tex]\[ \lim_{x \to \infty} \frac{x^3-2x+3}{x^2-5}. \][/tex]
Here, the degree of the numerator is 3 and the degree of the denominator is 2. Since the degree of the numerator is higher, this function does not have a horizontal asymptote.
2. Function [tex]\( v(x)=\frac{x^2-1}{x^2-5} \)[/tex]
For the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{x^2-1}{x^2-5}. \][/tex]
Since the degrees of the numerator and the denominator are the same, the horizontal asymptote is given by the ratio of the leading coefficients:
[tex]\[ \lim_{x \to \infty} \frac{x^2-1}{x^2-5} = \frac{1}{1} = 1. \][/tex]
Thus, this function [tex]\( v(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 1 \)[/tex].
3. Function [tex]\( g(x)=\frac{1-x}{x^2+2} \)[/tex]
For the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{1-x}{x^2+2}. \][/tex]
Here, the degree of the numerator is 1 and the degree of the denominator is 2. Since the degree of the denominator is higher, this function has a horizontal asymptote at [tex]\( y = 0 \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{1-x}{x^2+2} = 0. \][/tex]
Thus, [tex]\( g(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
4. Function [tex]\( w(x)=\frac{x+3 x^4}{x^2} \)[/tex]
For the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{x+3x^4}{x^2}. \][/tex]
Simplifying inside the limit:
[tex]\[ \lim_{x \to \infty} \frac{x(1 + 3x^3)}{x^2} = \lim_{x \to \infty} \frac{1 + 3x^3}{x} = \lim_{x \to \infty} (3x^2) \rightarrow \infty \text{ or } -\infty. \][/tex]
Since this limit grows without bound, [tex]\( w(x) \)[/tex] does not have a horizontal asymptote.
5. Function [tex]\( h(x)=\frac{x^3}{x^2-5 x^4} \)[/tex]
For the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{x^3}{x^2-5x^4}. \][/tex]
Simplifying inside the limit:
[tex]\[ \lim_{x \to \infty} \frac{x^3}{x^2(1-5x^2)} = \lim_{x \to \infty} \frac{x}{1-5x^2} = \lim_{x \to \infty} -\frac{1}{5x} \rightarrow 0. \][/tex]
Thus, [tex]\( h(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
Functions that have horizontal asymptotes are:
[tex]\[ v(x) = \frac{x^2-1}{x^2-5}, \quad g(x) = \frac{1-x}{x^2+2}, \quad \text{and} \quad h(x) = \frac{x^3}{x^2-5x^4}. \][/tex]
1. Function [tex]\( f(x)=\frac{x^3-2 x+3}{x^2-5} \)[/tex]
To find the horizontal asymptote, consider the degrees of the numerator and the denominator:
[tex]\[ \lim_{x \to \infty} \frac{x^3-2x+3}{x^2-5}. \][/tex]
Here, the degree of the numerator is 3 and the degree of the denominator is 2. Since the degree of the numerator is higher, this function does not have a horizontal asymptote.
2. Function [tex]\( v(x)=\frac{x^2-1}{x^2-5} \)[/tex]
For the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{x^2-1}{x^2-5}. \][/tex]
Since the degrees of the numerator and the denominator are the same, the horizontal asymptote is given by the ratio of the leading coefficients:
[tex]\[ \lim_{x \to \infty} \frac{x^2-1}{x^2-5} = \frac{1}{1} = 1. \][/tex]
Thus, this function [tex]\( v(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 1 \)[/tex].
3. Function [tex]\( g(x)=\frac{1-x}{x^2+2} \)[/tex]
For the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{1-x}{x^2+2}. \][/tex]
Here, the degree of the numerator is 1 and the degree of the denominator is 2. Since the degree of the denominator is higher, this function has a horizontal asymptote at [tex]\( y = 0 \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{1-x}{x^2+2} = 0. \][/tex]
Thus, [tex]\( g(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
4. Function [tex]\( w(x)=\frac{x+3 x^4}{x^2} \)[/tex]
For the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{x+3x^4}{x^2}. \][/tex]
Simplifying inside the limit:
[tex]\[ \lim_{x \to \infty} \frac{x(1 + 3x^3)}{x^2} = \lim_{x \to \infty} \frac{1 + 3x^3}{x} = \lim_{x \to \infty} (3x^2) \rightarrow \infty \text{ or } -\infty. \][/tex]
Since this limit grows without bound, [tex]\( w(x) \)[/tex] does not have a horizontal asymptote.
5. Function [tex]\( h(x)=\frac{x^3}{x^2-5 x^4} \)[/tex]
For the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{x^3}{x^2-5x^4}. \][/tex]
Simplifying inside the limit:
[tex]\[ \lim_{x \to \infty} \frac{x^3}{x^2(1-5x^2)} = \lim_{x \to \infty} \frac{x}{1-5x^2} = \lim_{x \to \infty} -\frac{1}{5x} \rightarrow 0. \][/tex]
Thus, [tex]\( h(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
Functions that have horizontal asymptotes are:
[tex]\[ v(x) = \frac{x^2-1}{x^2-5}, \quad g(x) = \frac{1-x}{x^2+2}, \quad \text{and} \quad h(x) = \frac{x^3}{x^2-5x^4}. \][/tex]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
