Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.1 - Finding Limits Using Tables and Graphs - Exercise Set - Page 1139: 29

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) $ f\left( -3 \right)=2$, because this point $\left( -3,2 \right)$ is shown by the solid dot in the provided graph. e) $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f=4$, since, as x approaches $-1$ from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $4$. f) $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=3$, since, as x approaches $-1$ from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 3. g) $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)$, does not exist, because the left hand and right-hand limits are unequal. h) $ f\left( -1 \right)$ does not exist, because this point is shown by the open dot in the provided graph. i) $\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2$, since, as x approaches $3$ from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $2$. j) $\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$, since, as x approaches $3$ from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of $2$. k) $\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=2$, because both the left hand and right-hand limits are equal to 2. l) $ f\left( 3 \right)=1$, because this point $\left( 3,1 \right)$ 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 solid 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 solid 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 solid 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 solid 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 solid dot in the provided graph with coordinates $\left( -3,2 \right)$. Thus, $ f\left( -3 \right)=2$ (e) Consider the provided limit $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$. To find $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=-1$ but from the left. As x approaches $-1$ from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 4. This point $\left( -1,4 \right)$ is shown by the open dot in the above graph. The point $\left( -1,4 \right)$ has a y-coordinate of 4. Thus, $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=4$. Hence, the value of $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 4. (f) Consider the provided limit $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$. To find $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=-1$ but from the right. As x approaches $-1$ from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 3. This point $\left( -1,3 \right)$ is shown by the open dot in the above graph. The point $\left( -1,3 \right)$ has a y-coordinate of 3. Thus $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=3$, . Hence, the value of $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is 3. (g) Consider the provided limit $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)$. To find $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=-1$ but from the right. As x approaches $-1$ from the right, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 3. This point $\left( -1,3 \right)$ is shown by the open dot in the above graph. The point $\left( -1,3 \right)$ has a y-coordinate of 3. Thus $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=3$, . Hence, the value of $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is 3. To find $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, examine the portion of the graph near $ x=-1$ but from the left. As x approaches $-1$ from the left, the value of $ f\left( x \right)$ gets closer to the y-coordinate of 4. This point $\left( -1,4 \right)$ is shown by the open dot in the above graph. The point $\left( -1,4 \right)$ has a y-coordinate of 4. Thus, $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=4$. Hence, the value of $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is 4. Since, $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=3$ and $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=4$. Here, both the left-hand limit and right-hand limit at $ x=-1$ are unequal, Hence, $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)$ does not exist. (h) Consider the provided function, $ f\left( -1 \right)$. To find $ f\left( -1 \right)$, examine the portion of the graph near $ x=-1$. The graph of “f” at $ x=-1$ is shown by the open dot in the provided graph with coordinates $\left( -1,3 \right)$. Thus, the function $ f\left( -1 \right)$ does not exist. (i) 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. (j) 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 $2$ . (k) 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 right hand and left-hand limit is the same at $ x=3$. Hence, $\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2$ (l) 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 solid dot in the provided graph with coordinates $\left( 3,1 \right)$. Thus, $ f\left( 3 \right)=1$
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