Answer
(a) $f(1)$ exists and equals $1$.
(b) $\lim_{x\to1}f(x)$ exists and equals $2$.
(c) $\lim_{x\to1}f(x)\ne f(1)$
(d) $f$ is discontinuous at $x=1$.
Work Step by Step
(a) From the graph, we can see right away from point $(1,1)$ that $f(1)$ exists and equals $1$.
(b) As $x$ approaches $1$ from both the right and the left, $f(x)$ approaches $2$. So $\lim_{x\to1}f(x)$ exists and equals $2$.
We can also do it algebraically:
- For $x\to1^-$, since $x\lt1$, we employ the function $f(x)=2x$
$$\lim_{x\to1^-}f(x)=\lim_{x\to1^-}2x=2\times1=2$$
- For $x\to1^+$, since $x\gt1$, we employ the function $f(x)=-2x+4$
$$\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(-2x+4)=-2\times1+4=2$$
As $\lim_{x\to1^+}f(x)=\lim_{x\to1^-}f(x)=2$, $\lim_{x\to1}f(x)$ exists and equals $2$.
(c) From the results in (a) and (b), $\lim_{x\to1}f(x)\ne f(1)$
(d) Because $\lim_{x\to1}f(x)\ne f(1)$, according to definition, $f$ is discontinuous at $x=1$.
