## Precalculus (6th Edition) Blitzer

a) $\underset{x\to -{{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-5$, since, as x approaches $-2$ from the left, the value of $f\left( x \right)$ gets closer to the y-coordinate of $-5$. b) $\underset{x\to -{{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=-1$, since, as x approaches $-2$ from the right, the value of $f\left( x \right)$ gets closer to the y-coordinate of $-1$. 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) $f\left( -2 \right)=-5$, because this point $\left( -2,-5 \right)$ 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 $-5$. This point $\left( -2,-5 \right)$ is shown by the solid dot in the above graph. The point $\left( -2,-5 \right)$ has a y-coordinate of $-5$. Thus, $\underset{x\to -{{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-5$. Hence, the value of $\underset{x\to -{{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $-5$. (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 $-1$. 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 $-1$ . (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 $-1$. 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 $-5$. This point $\left( -2,-5 \right)$ is shown by the solid dot in the above graph. The point $\left( -2,-5 \right)$ has a y-coordinate of $-5$. Thus, $\underset{x\to -{{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-5$. Hence, the value of $\underset{x\to -{{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $-5$. Since, $\underset{x\to -{{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=-1$ and $\underset{x\to -{{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-5$. 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 $\left( -2,-5 \right)$. Thus, $f\left( -2 \right)=-5$