## 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 $x\to-1$, 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 $x\to-1$, 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 and $1/(x+1)$ gets closer to $0$, meaning that the curve will approach the line $y=0$, which is also the horizontal asymptote. Also, because $1/(x+1)$ becomes closer to $0$, we can say the dominant term is $0$ as $x\to\pm\infty$. - As $x\to-1$, $(x+1)$ would approach $0$, meaning that the curve will approach $\infty$. So $x=-1$ is the vertical asymptote. And as $x\to-1$, $1/(x+1)$ becomes infinitely large. So the dominant term as $x\to-1$ is $1/(x+1)$. The graph is shown below. The red curve is the graph of $y=1/(x+1)$, while the blue line is the horizontal asymptote $y=0$ and the green one is the vertical asymptote $x=-1$.