University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Practice Exercises - Page 112: 56

Answer

The equations of the horizontal asymptotes of the functions are: a) $y=-1$ b) $y=1$ c) $y=-1$ and $y=1$ d) $y=1/3$

Work Step by Step

a) $$y=\frac{1-x^2}{x^2+1}$$ $$\lim_{x\to\pm\infty}y=\lim_{x\to\pm\infty}\frac{1-x^2}{x^2+1}$$ Divide both numerator and denominator by $x^2$: $$\lim_{x\to\pm\infty}y=\lim_{x\to\pm\infty}\frac{\frac{1}{x^2}-1}{1+\frac{1}{x^2}}=\frac{0-1}{1+0}=\frac{-1}{1}=-1$$ According to definition, $y=-1$ is the horizontal asymptote of the function. b) $$y=\frac{\sqrt x+4}{\sqrt{x+4}}$$ - Domain: $[0,\infty)$ So we only examine limits of $y$ as $x\to\infty$: $$\lim_{x\to\infty}y=\lim_{x\to\infty}\frac{\sqrt x+4}{\sqrt{x+4}}$$ Divide both numerator and denominator by $\sqrt x$: $$\lim_{x\to\infty}y=\lim_{x\to\infty}\frac{1+\frac{4}{\sqrt x}}{\frac{\sqrt{x+4}}{\sqrt x}}=\lim_{x\to\infty}\frac{1+\frac{4}{\sqrt x}}{\sqrt{1+\frac{4}{x}}}$$ $$\lim_{x\to\infty}y=\frac{1+0}{\sqrt{1+0}}=\frac{1}{\sqrt1}=1$$ According to definition, $y=1$ is the horizontal asymptote of the function. c) $$g(x)=\frac{\sqrt{x^2+4}}{x}$$ $$\lim_{x\to\pm\infty}g(x)=\lim_{x\to\pm\infty}\frac{\sqrt{x^2+4}}{x}$$ - For $x\to\infty$: we assume $x\gt0$, and $x=|x|=\sqrt{x^2}$ $$\lim_{x\to\infty}g(x)=\lim_{x\to\infty}\frac{\sqrt{x^2+4}}{\sqrt{x^2}}=\lim_{x\to\infty}\sqrt{1+\frac{4}{x^2}}=\sqrt{1+0}=1$$ - For $x\to-\infty$: we assume $x\lt0$, and $x=-|x|=-\sqrt{x^2}$ $$\lim_{x\to-\infty}g(x)=\lim_{x\to-\infty}\frac{\sqrt{x^2+4}}{-\sqrt{x^2}}=\lim_{x\to-\infty}-\sqrt{1+\frac{4}{x^2}}=-\sqrt{1+0}=-1$$ According to definition, $y=1$ and $y=-1$ are the horizontal asymptotes of the function. d) $$y=\sqrt{\frac{x^2+9}{9x^2+1}}$$ $$\lim_{x\to\pm\infty}y=\lim_{x\to\pm\infty}\sqrt{\frac{x^2+9}{9x^2+1}}=\sqrt{\lim_{x\to\pm\infty}\frac{x^2+9}{9x^2+1}}$$ Divide both numerator and denominator by $x^2$: $$\lim_{x\to\pm\infty}y=\sqrt{\lim_{x\to\pm\infty}\frac{1+\frac{9}{x^2}}{9+\frac{1}{x^2}}}=\sqrt{\frac{1+0}{9+0}}=\sqrt{\frac{1}{9}}=\frac{1}{3}$$ According to definition, $y=1/3$ is the horizontal asymptote of the function.
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