Answer
(a) $\lim_{x\to2^+}=\lim_{x\to2^-}=1$ and $f(2)=2$
(b) $\lim_{x\to2}$ does exist and equals $1$.
(c) $\lim_{x\to-1^+}=\lim_{x\to-1^-}=4$
(d) $\lim_{x\to-1}$ does exist and equals $4$
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 $1$, too.
And we can check algebraically: $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(x/2)=2/2=1$
- For $x=2$, $f(x)=2$. Therefore, $f(2)=2$
(b) $\lim_{x\to2}f(x)$ exists and equals $1$ since $\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=1$, so $f(x)$ would gets arbitrarily close to $1$ as $x$ approaches $2$ from either side.
(c) Find $\lim_{x\to-1^+}f(x)$ and $\lim_{x\to-1^-}f(x)$
- From the graph, we can see as $x$ approaches $-1$ from the left, $f(x)$ gets arbitrarily close to $4$.
And we can check algebraically: $\lim_{x\to-1^-}f(x)=\lim_{x\to-1^-}(3-x)=3-(-1)=4$
- From the graph, we can also see as $x$ approaches $-1$ from the right, $f(x)$ gets arbitrarily close to $4$.
And we can check algebraically: $\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}(3-x)=3-(-1)=4$
(d) $\lim_{x\to-1}f(x)$ exists and equals $4$ since $\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=4$, so $f(x)$ would get arbitrarily close to $4$, too, as $x$ approaches $-1$ from either side.
