Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises - Page 71: 48

Answer

(a) (i) $1$; (ii) $-1$; (iii) $DNE$; (iv) $-1$; (v) $1$; (vi) $DNE$ (b) $a=k\pi$, where $k$ is integer (c) see graph

Work Step by Step

Recall the sign function: $$\begin{align*} sgn(x)=\begin{cases} -1\text{ for }x<0,\\ 0,\text{ for }x=0\\ +1\text{ for }x>0 \end{cases} \end{align*}$$ $g(x)=sgn(\sin(x))$ (a) Evaluate the limits: (i)$\lim\limits_{x \to 0^{+}}g(x)=1$ because as $x$ approaches $0$ from the right, $\sin(x)$ gets closer to but always greater than $0$, so the sign is positive and the limit is positive $1$. (ii)$\lim\limits_{x \to 0^{-}}g(x)=-1$ because as x approaches $0$ from the left, $\sin(x)$ gets closer to but always less than $0$, so the sign is negative and the limit is negative $1$ (iii)$\lim\limits_{x \to 0}g(x)=DNE$ because the left and right limits are not the same. (iv)$\lim\limits_{x \to \pi^{+}}g(x)=-1$ because as $x$ approaches $\pi$ from the right, $\sin(x)$ gets closer to but always less than $0$, so the sign is negative and the limit is negative $1$ (v)$\lim\limits_{x \to \pi^{-}}g(x)=1$ because as $x$ approaches $\pi$ from the left, $\sin(x)$ gets closer to but always greater than $0$, so the sign is positive and the limit is positive $1$ (vi)$\lim\limits_{x \to \pi}g(x)=DNE$ because the left and right limits are not the same. (b) $\lim\limits_{x \to a}g(x)=DNE$ when a is a multiple of $\pi$ (c) see graph
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