Answer
a) $\underset{x\to -{{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2$, since, as x approaches $-3$ from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $2$.
b) $\underset{x\to -{{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$, since, as x approaches $-3$ from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $2$.
c) $\underset{x\to -3}{\mathop{\lim }}\,f\left( x \right)=2$, because both the left hand and right-hand limits are equal to 2.
d)
The function $ f\left( -3 \right)$ does not exist, because this point $\left( -3,2 \right)$ is shown by the open dot in the provided graph.
e) $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$, since, as x approaches 0 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 0.
f) $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=3$, since, as x approaches 0 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 3.
g)
$\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$, does not exist, because the left hand and right-hand limits are unequal.
h) $ f\left( 0 \right)$ does not exist, because this point is shown by the open dot in the provided graph.
i) $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1$, since, as x approaches 2 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 1.
j) $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=1$, since, as x approaches 2 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 1.
k) $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=1$, because both the left hand and right-hand limits are equal to 1.
l)
$ f\left( 2 \right)=-1$, because this point $\left( 2,-1 \right)$ is shown by the solid dot in the provided graph.
Work Step by Step
(a)
Consider the provided limit $\underset{x\to -{{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to -{{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=-3$ but from the left.
As x approaches $-3$ from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $2$. This point $\left( -3,2 \right)$ is shown by the open dot in the above graph.
The point $\left( -3,2 \right)$ has a y-coordinate of 2.
Thus, $\underset{x\to -{{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to -{{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 2.
(b)
Consider the provided limit $\underset{x\to -{{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to -{{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=-3$ but from the right.
As x approaches $-3$ from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $2$. This point $\left( -3,2 \right)$ is shown by the open dot in the above graph.
The point $\left( -3,2 \right)$ has a y-coordinate of $2$.
Thus, $\underset{x\to -{{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to -{{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $2$ .
(c)
Consider the provided limit $\underset{x\to -3}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to -{{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=-3$ but from the right.
As x approaches $-3$ from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $2$. This point $\left( -3,2 \right)$ is shown by the open dot in the above graph.
The point $\left( -3,2 \right)$ has a y-coordinate of $2$.
Thus, $\underset{x\to -{{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to -{{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $2$.
To find $\underset{x\to -{{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=-3$ but from the left.
As x approaches $-3$ from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $2$. This point $\left( -3,2 \right)$ is shown by the open dot in the above graph.
The point $\left( -3,2 \right)$ has a y-coordinate of $2$.
Thus, $\underset{x\to -{{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to -{{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $2$.
Since, $\underset{x\to -{{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$ and $\underset{x\to -{{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Here, both the left-hand limit and right-hand limit at $ x=-3$ are equal,
Hence, $\underset{x\to -3}{\mathop{\lim }}\,f\left( x \right)=2$.
(d)
Consider the provided function, $ f\left( -3 \right)$.
To find $ f\left( -3 \right)$, examine the portion of the graph near $ x=-3$.
The graph of “f” at $ x=-3$ is shown by the open dot in the provided graph with coordinates $\left( -3,2 \right)$.
Thus, the function $ f\left( -3 \right)$ does not exist.
(e)
Consider the provided limit $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=0$ but from the left.
As x approaches 0 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 0. This point $\left( 0,0 \right)$ is shown by the open dot in the above graph.
The point $\left( 0,0 \right)$ has a y-coordinate of 0.
Thus, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$.
Hence, the value of $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 0.
(f)
Consider the provided limit $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=0$ but from the right.
As x approaches 0 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 3. This point $\left( 0,3 \right)$ is shown by the solid dot in the above graph.
The point $\left( 0,3 \right)$ has a y-coordinate of 3.
Thus $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$.
Hence, the value of $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is 3.
(g)
Consider the provided limit $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=0$ but from the right.
As x approaches 0 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 3. This point $\left( 0,3 \right)$ is shown by the solid dot in the above graph.
The point $\left( 0,3 \right)$ has a y-coordinate of 3.
Thus $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=3$.
Hence, the value of $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is 3.
To find $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=0$ but from the left.
As x approaches 0 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 0. This point $\left( 0,0 \right)$ is shown by the open dot in the above graph.
The point $\left( 0,0 \right)$ has a y-coordinate of 0.
Thus, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$.
Hence, the value of $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 0.
Since, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=3$ and $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$.
Here, both the left-hand limit and right-hand limit at $ x=0$ are unequal,
Hence, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ does not exist.
(h)
To find $ f\left( 0 \right)$, examine the portion of the graph near $ x=0$.
The graph of “f” at $ x=0$ is shown by the open dot in the provided graph with coordinates $\left( 0,0 \right)$.
Thus, the function $ f\left( 0 \right)$ does not exist.
(i)
Consider the provided limit $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=2$ but from the left.
As x approaches 2 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 1. This point $\left( 2,1 \right)$ is shown by the open dot in the above graph.
The point $\left( 2,1 \right)$ has a y-coordinate of 1.
Thus, $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1$.
Hence, the value of $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 1.
(j)
To find $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=2$ but from the right.
As x approaches 2 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 1. This point $\left( 2,1 \right)$ is shown by the open dot in the above graph.
The point $\left( 2,1 \right)$ has a y-coordinate of 1.
Thus, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=1$.
Hence, the value of $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is 1.
(k)
Consider the provided limit $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=2$ but from the right.
As x approaches 2 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 1. This point $\left( 2,1 \right)$ is shown by the open dot in the above graph.
The point $\left( 2,1 \right)$ has a y-coordinate of 1.
Thus, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=1$.
Hence, the value of $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is 1.
.
To find $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=2$ but from the left.
As x approaches 2 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 1. This point $\left( 2,1 \right)$ is shown by the open dot in the above graph.
The point $\left( 2,1 \right)$ has a y-coordinate of 1.
Thus, $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1$.
Hence, the value of $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 1.
Since, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=1$ and $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1$, here both the right hand and left-hand limit is the same at $ x=2$.
Hence, $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=1$
(l)
Consider the provided function, $ f\left( 2 \right)$.
To find $ f\left( 2 \right)$, examine the portion of the graph near $ x=2$.
The graph of “f” at $ x=2$ is shown by the solid dot in the provided graph with coordinates $\left( 2,-1 \right)$.
Thus, $ f\left( 2 \right)=-1$
