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: 6


(a) $f(1)$ exists and equals $1$. (b) $\lim_{x\to1}f(x)$ exists and equals $2$. (c) $\lim_{x\to1}f(x)\ne f(1)$ (d) $f$ is discontinuous at $x=1$.

Work Step by Step

(a) From the graph, we can see right away from point $(1,1)$ that $f(1)$ exists and equals $1$. (b) As $x$ approaches $1$ from both the right and the left, $f(x)$ approaches $2$. So $\lim_{x\to1}f(x)$ exists and equals $2$. We can also do it algebraically: - For $x\to1^-$, since $x\lt1$, we employ the function $f(x)=2x$ $$\lim_{x\to1^-}f(x)=\lim_{x\to1^-}2x=2\times1=2$$ - For $x\to1^+$, since $x\gt1$, we employ the function $f(x)=-2x+4$ $$\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(-2x+4)=-2\times1+4=2$$ As $\lim_{x\to1^+}f(x)=\lim_{x\to1^-}f(x)=2$, $\lim_{x\to1}f(x)$ exists and equals $2$. (c) From the results in (a) and (b), $\lim_{x\to1}f(x)\ne f(1)$ (d) Because $\lim_{x\to1}f(x)\ne f(1)$, according to definition, $f$ is discontinuous at $x=1$.
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