University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 68: 78


(a) We cannot conclude anything about the values of $f, g, h$ at $x=2$. (b) $f(2)=0$ is possible. (c) $\lim_{x\to2}f(x)=0$ is not possible.

Work Step by Step

$g(x)\le f(x)\le h(x)$ for all $x\ne2$ $\lim_{x\to2}g(x)=\lim_{x\to2}h(x)=-5$ According to the Sandwich Theorem, this means $\lim_{x\to2}f(x)=-5$ (a) Can we conclude anything about the values of $f, g, h$ at $x=2$? Unfortunately, no. Recall that when we are given or need to find the limit of a function $f(x)$ as $x\to c$ (or $x\to2$ in this case), we measure the value that $f(x)$ would reach as $x$ APPROACHES $c$. So we do not actually work with the value of $f(x)$ as $x=c$; hence, it is not necessarily true that $f(c)=\lim_{x\to c}f(x)$ (a lot of cases have been shown to be wrong in this section). Therefore, we cannot conclude anything about the values of $f, g, h$ at $x=2$. They might equal $-5$, or they might not. (b) Could $f(2)=0$? Yes, it is possible. Or $f(2)=1$, $f(2)=2$, $f(2)=-98/5$, etc. any of them would also be possible. (c) Could $\lim_{x\to2}f(x)=0$? No. As proved according to the Sandwich Theorem above, $\lim_{x\to2}f(x)=-5$ Notice that the inequality $g(x)\le f(x)\le h(x)$ does not have to be true for $x=2$, for reasons, again, that we do not work with $x=2$ when we work with limit as $x\to2$.
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