University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.5 - Continuity - Exercises - Page 94: 16

Answer

$y=\frac{x+3}{(x+2)(x-5)}$ is continuous as $x\in(-\infty,-2)\cup(-2,5)\cup(5,\infty)$

Work Step by Step

$$y=\frac{x+3}{x^2-3x-10}=\frac{x+3}{(x+2)(x-5)}$$ - Domain: $(-\infty,-2)\cup(-2,5)\cup(5,\infty)$ We would find all the points where the function is discontinous first, then all the remaining points would be where the function is continous. - At $x=-2$ and $x=5$: $$\lim_{x\to-2}y=\lim_{x\to-2}\frac{x+3}{(x+2)(x-5)}$$ $$\lim_{x\to5}y=\lim_{x\to5}\frac{x+3}{(x+2)(x-5)}$$ As $x\to-2$, $(x+2)\to0$, meaning that $(x+2)(x-5)$ will approach $0$ and $\frac{x+3}{(x+2)(x-5)}$, hence, will approach infinity, not any single value. Similarly, as $x\to5$, $(x-5)\to0$, meaning that $(x+2)(x-5)$ will approach $0$ and $\frac{x+3}{(x+2)(x-5)}$, hence, will approach infinity again. In other words, $\lim_{x\to-2}\frac{x+3}{(x+2)(x-5)}$ and $\lim_{x\to5}\frac{x+3}{(x+2)(x-5)}$ do not exist. So the function is discontinous at $x=-2$ and $x=5$. Therefore, $y=\frac{x+3}{(x+2)(x-5)}$ is continous in its domain, $(-\infty,-2)\cup(-2,5)\cup(5,\infty)$
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