Answer
(a) False
(b) False
(c) True
(d) True
(e) True
Work Step by Step
a) The limit $\lim\limits_{x \to 2} f(x) f(x)$ does not exist the reason being when value of $f(2)$ jumps to $0$, for values of $x to 2$ from the left side the $f(x) to 1$, and for $x$ will approach to $x=2$ from the right, so, $f(x) \to 1$ Thus, when $x\to2$, $f(x) \to1$, that is, $\lim\limits_{x \to 2} f(x)=1$
b)As explained in part (a), so $\lim\limits_{x \to 2} f(x)=1 \to False$
c) $\lim_{x\to1}f(x)$ does not exist \to TRUE.
as $f(x)$ jumps at $x=1$. For values of $x \to1$ from the left side, $f(x)\to -2$, and for $x \to 1$ from the right side, the $f(x) \to 0$.
Thus, there can be no single value that $f(x)$ approaches when $x\to1$.
d) $\lim\limits_{x\to c}f(x)$ exists at every point $c$ in $(-1,1)$
The given statement is is true. Because the graph as $x$ in $(-1,1)$ is continuous, and there has only one corresponding value of $y$, this shows that all the limits exists here.
e) $\lim\limits_{x\to c}f(x)$ exists at every point $c$ in $(1,3)$
The given statement is True. As $x$ in $(1,2)$ and $(2,3)$, as shown in part $(d)$, the graph is continuous and one-to-one, so all the limits exists.
At $x=2$, as shown in part $(a)$, the limit does exist and equal to $1$.
Hence, $\lim\limits_{x\to c}f(x)$ exists at every point c on $(1,3)$
