## University Calculus: Early Transcendentals (3rd Edition)

- Horizontal asymptote: $y=0$ - Vertical asymptote: $x=1$ As $x\to\pm\infty$, the dominant term is $0$. As $x\to1$, the dominant term is $1/(x-1)$.
$$y=\frac{1}{x-1}$$ We are interested in the behavior of function $y$ as $x\to\pm\infty$ as well as the behavior of $y$ as $x\to1$, which is where the denominator is $0$. We can rewrite the function into a polynomial with a remainder as follows: $$y=\frac{1}{x-1}=0+\frac{1}{x-1}$$ - As $x\to\pm\infty$, $(x-1)$ would get infinitely large, meaning that the curve will approach the line $y=0$, which is also the horizontal asymptote. Also, as $x$ becomes infinitely large, $1/(x-1)$ will become closer to $0$. So we can say the dominant term is $0$ as $x\to\pm\infty$. - As $x\to1$, $(x-1)$ would approach $0$, meaning that the curve will approach $\infty$. So $x=1$ is the vertical asymptote. And as $x\to1$, $1/(x-1)$ becomes infinitely large. So the dominant term as $x\to1$ is $1/(x-1)$. The graph is shown below. The red curve is the graph of $y$, while the blue line is the horizontal asymptote $y=0$ and the green one is the vertical asymptote $x=1$.