University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 108: 68


- Horizontal asymptote: $y=2$ - Vertical asymptote $x=-1$ As $x\to\pm\infty$, the dominant term is $2$. As $x\to-1$, the dominant term is $-2/(x+1)$.

Work Step by Step

$$y=\frac{2x}{x+1}$$ 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: $$y=\frac{2x}{x+1}=\frac{(2x+2)-2}{x+1}=2-\frac{2}{x+1}$$ - As $x\to\pm\infty$, $(x+1)$ approaches $\pm\infty$ as well and $2/(x+1)$ gets closer to $0$. Therefore, the function $y=2-\frac{2}{x+1}$ will approach $2$, meaning that the curve will approach the line $y=2$, which is also the horizontal asymptote. One more thing, because $2/(x+1)$ becomes closer to $0$, $2$ will dominate when $x$ is numerically large. We can say the dominant term is $2$ as $x\to\pm\infty$. - As $x\to-1$, $(x+1)$ will approach $0$ and $-2/(x+1)$ will approach $-\infty$. Therefore, the function $y=2-\frac{2}{x+1}$ will approach $-\infty$, meaning that the curve will approach $-\infty$ as well. So $x=-1$ is the vertical asymptote. And since $-2/(x+1)$ approaches $-\infty$, or we might say becomes infinitely large in absolute value, as $x\to-1$, the dominant term is $-2/(x+1)$ here. The graph is shown below. The red curve is the graph of $y=\frac{2x}{x+1}$, while the blue line is the horizontal asymptote $y=2$ and the green one is the vertical asymptote $x=-1$. The purple curve is the dominant term $-2/(x+1)$ as $x\to-1$.
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