Answer:
vertical asymptote. x = 5
horizontal asymptote, y = 1
Step-by-step explanation:
The vertical asymptote of f(x)= x/x-5 is gotten when the denominator x - 5 = 0 ⇒ x = 5.
The horizontal asymptote of f(x)= x/x-5 is gotten when we find [tex]\lim_{x \to \infty} f(x)[/tex].
So
[tex]\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{x}{x - 5} \\= \lim_{x \to +\infty} \frac{1}{1 - \frac{5}{x} } \\= \frac{1}{1 - \frac{5}{+\infty }} \\= \frac{1}{1 - 0} \\= \frac{1}{1} \\= 1[/tex]
[tex]\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{x}{x - 5} \\= \lim_{x \to -\infty} \frac{1}{1 - \frac{5}{x} } \\= \frac{1}{1 - \frac{5}{-\infty }} \\= \frac{1}{1 + 0} \\= \frac{1}{1} \\= 1[/tex]
Since
[tex]\lim_{n \to +\infty} f(x) = \lim_{n \to -\infty} f(x) = 1. The limit exists\\ horizontal asymptote = \lim_{n \to \infty} f(x) = 1[/tex]
