Answer :
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]
