University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Practice Exercises - Page 110: 1

Answer

The graph is enclosed below. 1) At $x=-1$: $\lim_{x\to-1}f(x)=\lim_{x\to-1^+}f(x)=\lim_{x\to-1^-}f(x)=1$ $f(x)$ is left-continuous, right-continuous, and overall, continuous at $x=-1$. 2) At $x=0$: $\lim_{x\to0}f(x)=\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=0$ $f(x)$ is discontinuous at $x=0$. It is also not left-continuous or right-continuous at $x=0$. This discontinuity is removable. 3) At $x=1$: $\lim_{x\to1^+}f(x)=1$, $\lim_{x\to1^-}f(x)=-1$, $\lim_{x\to1}f(x)$ does not exist. $f(x)$ is discontinuous at $x=1$. But it is right-continuous here. This discontinuity is non-removable.
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Work Step by Step

The graph is enclosed below. 1) At $x=-1$: - $\lim_{x\to-1^-}f(x)=\lim_{x\to-1^-}1=1$ - $\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}-x=-(-1)=1$ So $\lim_{x\to-1}f(x)=\lim_{x\to-1^+}f(x)=\lim_{x\to-1^-}f(x)=1$ - $f(-1)=1$ So $\lim_{x\to-1}f(x)=\lim_{x\to-1^+}f(x)=\lim_{x\to-1^-}f(x)=f(-1)=1$ Therefore, $f(x)$ is left-continuous, right-continuous, and overall, continuous at $x=-1$. 2) At $x=0$: - $\lim_{x\to0^-}f(x)=\lim_{x\to0^-}-x=-0=0$ - $\lim_{x\to0^+}f(x)=\lim_{x\to0^+}-x=-0=0$ So $\lim_{x\to0}f(x)=\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=0$ - $f(0)=1$ So $\lim_{x\to0}f(x)\ne f(0)$. Neither is $f(0)$ equal with $\lim_{x\to0^+}f(x)$ or $\lim_{x\to0^-}f(x)$ Therefore, $f(x)$ is discontinuous at $x=0$. It is also not left-continuous or right-continuous at $x=0$. This discontinuity is removable because only $\lim_{x\to0}f(x)\ne f(0)$, so we can extend the function $f(x)$ to cover $f(0)=0$ so that $\lim_{x\to0}f(x)=f(0)$, thus removing the discontinuity. 3) At $x=1$: - $\lim_{x\to1^-}f(x)=\lim_{x\to1^-}-x=-1$ - $\lim_{x\to1^+}f(x)=\lim_{x\to1^+}1=1$ So $\lim_{x\to1^+}f(x)\ne\lim_{x\to1^-}f(x)$. $\lim_{x\to1}f(x)$, as a result, does not exist. - $f(1)=1$ Since $\lim_{x\to1}f(x)$ does not exist, $f(x)$ is discontinuous at $x=1$. However, since $\lim_{x\to1^+}f(x)=f(1)=1$, $f(x)$ is right-continuous at $x=1$. This discontinuity is not removable because $\lim_{x\to1}f(x)$ does not exist. We cannot remove the discontinuity by making $f(x)$ include any values so that $\lim_{x\to1}f(x)=f(1)$.
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