## University Calculus: Early Transcendentals (3rd Edition)

(a) Domain: $(-\infty,\infty)$ - Range: $[-1,1]$ (b) At $c\in(-\infty,-1)\cup(-1,1)\cup(1,\infty)$, $\lim_{x\to c}f(x)$ exists. (c) and (d) There are no points at which either only the left-hand or only the right-hand limit exists.
(a) The domain, where $x$ is defined, has been stated in the function, which is $(-\infty,\infty)$. The range: Looking at the graph, we notice that - $f(x)\ge-1$. The lowest point is at $f(x)=-1$, when $x=-1$. - $f(x)\le1$. The highest point is at $f(x)=1$, when $x=0$ and $x=1$ So the range is $[-1,1]$. (b) $\lim_{x\to c}f(x)$ exists when as $x$ approaches $c$ from either the left or the right, $f(x)$ would still approach the same value. Most points in continuous graphs would have $\lim_{x\to c}f(x)$ exist, except some special cases. We see that as $x=c\in(-\infty,-1)\cup(-1,1)\cup(1,\infty)$, as $x$ approaches an arbitrary $c$ from either left or right, $f(x)$ would still approach the same value. Therefore, at $c\in(-\infty,-1)\cup(-1,1)\cup(1,\infty)$, $\lim_{x\to c}f(x)$ exists. - At $c=-1$, $\lim_{x\to-1^+}f(x)=-1$, while $\lim_{x\to-1^-}f(x)=0$, so $\lim_{x\to-1}f(x)$ does not exist. - At $c=1$, $\lim_{x\to1^+}f(x)=0$, while $\lim_{x\to1^-}f(x)=1$, so $\lim_{x\to1}f(x)$ does not exist. (c) Examining the whole domain, we conclude that there are no points at which only the left-hand limit exists. (d) Examining the whole domain, we conclude that there are no points at which only the right-hand limit exists.