## Precalculus (6th Edition) Blitzer

a) $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=4$, since, as x approaches 2 from the left the value of $f\left( x \right)$ gets closer to the y-coordinate of 4. 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 the left hand and right-hand limits are unequal. d) $f\left( 2 \right)=4$, because this point $\left( 2,4 \right)$ is shown by the solid dot in the provided graph.
(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 4. This point $\left( 2,4 \right)$ is shown by the solid dot in the above graph. The point $\left( 2,4 \right)$ has a y-coordinate of 4. Thus, $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=4$. Hence, the value of $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 4. (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 open 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 open 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 4. This point $\left( 2,4 \right)$ is shown by the solid dot in the above graph. The point $\left( 2,4 \right)$ has a y-coordinate of 4. Thus, $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=4$. Hence, the value of $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 4. Since, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$ and $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=4$. Here, both the left-hand limit and right-hand limit at $x=2$ are unequal. Hence the provided limit $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ 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 $\left( 2,4 \right)$. Thus, $f\left( 2 \right)=4$