Answer
a)
$\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1$, since, as x approaches 2 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 1.
b) $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$, since, as x approaches 2 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $2$.
c) $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$, does not exist, because both the left hand and right-hand limits are unequal at $ x=2$.
d) $ f\left( 2 \right)=2$, because this point $\left( 2,2 \right)$ is shown by the solid dot in the provided graph.
e) $\underset{x\to {{2.5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2$, since, as x approaches 2.5 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 2.
f) $\underset{x\to {{2.5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$, since, as x approaches 2.5 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 2.
g)
$\underset{x\to 2.5}{\mathop{\lim }}\,f\left( x \right)=2$, because the left hand and right-hand limits at $ x=2.5$ are equal.
h) $ f\left( 2.5 \right)=2$, because this point is shown by the solid dot in the provided graph.
Work Step by Step
(a)
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.
(b)
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 $2$. This point $\left( 2,2 \right)$ is shown by the solid dot in the above graph.
The point $\left( 2,2 \right)$ has a y-coordinate of $2$.
Thus, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $2$ .
(c)
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 $2$. This point $\left( 2,2 \right)$ is shown by the solid dot in the above graph.
The point $\left( 2,2 \right)$ has a y-coordinate of $2$.
Thus, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $2$ .
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, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$ and $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1$.
Here, both the left-hand limit and right-hand limit at $ x=2$ are unequal,
Hence, $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ does not exist.
(d)
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 $\left( 2,2 \right)$.
Thus, $ f\left( 2 \right)=2$.
(e)
To find $\underset{x\to {{2.5}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=2.5$ but from the left.
As x approaches 2.5 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 2. This point $\left( 2.5,2 \right)$ is on the parallel line shown at $ y=2$ in the above graph.
The point $\left( 2.5,2 \right)$ has a y-coordinate of 2.
Thus, $\underset{x\to {{2.5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to {{2.5}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 2.
(f)
Consider the provided limit $\underset{x\to {{2.5}^{+}}}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to {{2.5}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=2.5$ but from the right.
As x approaches 2.5 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 2. This point $\left( 2.5,2 \right)$ is on the parallel line shown at $ y=2$ in the above graph
Thus $\underset{x\to {{2.5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to {{2.5}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is 2.
(g)
Consider the provided limit $\underset{x\to 2.5}{\mathop{\lim }}\,f\left( x \right)$.
To find $\underset{x\to {{2.5}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=2.5$ but from the right.
As x approaches 2.5 from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 2. This point $\left( 2.5,2 \right)$ is on the parallel line shown at $ y=2$ in the above graph
Thus $\underset{x\to {{2.5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to {{2.5}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is 2.
To find $\underset{x\to {{2.5}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=2.5$ but from the left.
As x approaches 2.5 from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 2. This point $\left( 2.5,2 \right)$ is on the parallel line shown at $ y=2$ in the above graph.
The point $\left( 2.5,2 \right)$ has a y-coordinate of 2.
Thus, $\underset{x\to {{2.5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Hence, the value of $\underset{x\to {{2.5}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 2.
Since, $\underset{x\to {{2.5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$ and $\underset{x\to {{2.5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2$.
Here, both the left-hand limit and right-hand limit at $ x=2.5$ are equal,
Hence, $\underset{x\to 2.5}{\mathop{\lim }}\,f\left( x \right)=2$ .
(h)
Consider the provided function, $ f\left( 2.5 \right)$.
To find $ f\left( 2.5 \right)$, examine the portion of the graph near $ x=2.5$.
The graph of “f” at $ x=2.5$ is shown by the solid dot in the provided graph with coordinates $\left( 2.5,2 \right)$.
Thus, the function $ f\left( 2.5 \right)=2$.
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