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 95: 60



Work Step by Step

$$f(x)=\frac{x+6}{x+1}$$ This function is continuous for all $x$ except $x=-1$, where it has a nonremovable discontinuity. The reason is stated as follows: 1) For all $x\ne-1$, we have $\lim_{x\to c}\frac{x+6}{x+1}=\frac{c+6}{c+1}=f(c)$, meaning that $f$ is continuous for all $x\ne-1$. - As $x\to-1$, though, $(x+1)$ approaches $0$, so $\frac{x+6}{x+1}$ will approach $\infty$. This means $\lim_{x\to-1}f(x)$ does not exist. $f(x)$ is discontinuous at $x=-1$ as a result. 2) This discontinuity is a nonremovable one, because both $\lim_{x\to-1}f(x)$ does not exist. Therefore, there is no way we can extend $f(x)$ to include a value of $f(-1)=\lim_{x\to-1}f(x)$ in order to remove the discontinuity.
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