#### Answer

(a) $\lim_{x\to2^+}f(x)=2$ and $\lim_{x\to2^-}f(x)=1$
(b) $\lim_{x\to2}f(x)$ does not exist.
(c) $\lim_{x\to4^+}f(x)=3$ and $\lim_{x\to4^-}f(x)=3$
(d) $\lim_{x\to4}f(x)$ exists and equals $3$.

#### Work Step by Step

(a) Find $\lim_{x\to2^+}f(x)$ and $\lim_{x\to2^-}f(x)$
- From the graph, we can see as $x$ approaches $2$ from the left, $f(x)$ gets arbitrarily close to $1$.
And we can check algebraically: $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(3-x)=3-2=1$
- From the graph, we can also see as $x$ approaches $2$ from the right, $f(x)$ gets arbitrarily close to $2$.
And we can check algebraically: $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(x/2+1)=(2/2)+1=2$
(b) $\lim_{x\to2}f(x)$ does not exist since $\lim_{x\to2^-}f(x)\ne\lim_{x\to2^+}f(x)$, so there is no single value that $f(x)$ would reach as $x$ approaches $2$.
(c) Find $\lim_{x\to4^+}f(x)$ and $\lim_{x\to4^-}f(x)$
- From the graph, we can see as $x$ approaches $4$ from the left, $f(x)$ gets arbitrarily close to $3$.
And we can check algebraically: $\lim_{x\to4^-}f(x)=\lim_{x\to4^-}(x/2+1)=(4/2)+1=3$
- From the graph, we can also see as $x$ approaches $4$ from the right, $f(x)$ gets arbitrarily close to $3$.
And we can check algebraically: $\lim_{x\to4^+}f(x)=\lim_{x\to4^+}(x/2+1)=(4/2)+1=3$
(d) $\lim_{x\to4}f(x)$ exists and equals $3$ since $\lim_{x\to4^-}f(x)=\lim_{x\to4^+}f(x)=3$, so there is this value of $3$ that $f(x)$ would reach as $x$ approaches $4$ from either side.