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: 13

Answer

$y=\frac{1}{x-2}-3x$ is continous as $x\in(-\infty,2)\cup(2,\infty)$

Work Step by Step

$$y=\frac{1}{x-2}-3x$$ - Domain: $(-\infty,2)\cup(2,\infty)$ We would find all the points where the function is discontinous first, then all the remaining points would be where the function is continous. We see that at $x=2$: $$\lim_{x\to2}y=\lim_{x\to2}\Big(\frac{1}{x-2}-3x\Big)$$ As $x\to2$, $(x-2)\to0$, meaning that $\frac{1}{x-2}$ will approach infinity, not any single value. In other words, $\lim_{x\to2}\Big(\frac{1}{x-2}-3x\Big)$ does not exist. So the function is discontinous at $x=2$. That is also the only point where $y$ is discontinous. Therefore, $y=\frac{1}{x-2}-3x$ is continous in its domain, $(-\infty,2)\cup(2,\infty)$
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