University Calculus: Early Transcendentals (3rd Edition)

(a) Domain: $[0,2]$ - Range: $(0,1]$ (b) At every point $c\in(0,1)\cup(1,2)$, $\lim_{x\to c}f(x)$ exists. (c) At $c=2$, only the left-hand limit exists. (d) At $c=0$, only the right-hand limit exists.
(a) The domain, where $x$ is defined, has been stated in the function, which is $[0,2]$. The range: Looking at the graph, we notice that - $f(x)\gt0$. $f(x)$ does not reach down until $0$ at $x=1$, because at $x=1$, the graph turns to the purple line, and $f(x)=1$. - $f(x)\le1$. There is no higher point than $f(x)=1$. So the range is $(0,1]$. (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 $x=c\in(0,1)\cup(1,2)$, as $x$ approaches an arbitrary $c$ from either left or right, $f(x)$ would still approach the same value. Therefore, at $c\in(0,1)\cup(1,2)$, $\lim_{x\to c}f(x)$ exists. - At $c=0$, $\lim_{x\to0^-}f(x)$ does not exist, so $\lim_{x\to0}f(x)$ does not exist. - At $c=2$, $\lim_{x\to2^+}f(x)$ does not exist, so $\lim_{x\to2}f(x)$ does not exist. - At $c=1$, $\lim_{x\to1^+}f(x)=1$, while $\lim_{x\to1^-}f(x)=0$, so $\lim_{x\to1}f(x)$ does not exist. (c) As mentioned in (b), at $c=2$, the right-hand limit does not exist, only the left-hand one does. (d) As mentioned in (b) again, at $c=0$, the left-hand limit does not exist, only the right-hand one does.