Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises - Page 92: 6

Answer

(a) $\lim\limits_{x \to -3^{-}} h(x) = 4$ (b) $\lim\limits_{x \to -3^{+}} h(x) = 4$ (c) $\lim\limits_{x \to -3} h(x) = 4$ (d) $h(-3)$ does not exist (e) $\lim\limits_{x \to 0^{-}} h(x) = 1$ (f) $\lim\limits_{x \to 0^{+}} h(x) = -1$ (g) $\lim\limits_{x \to 0} h(x)$ does not exist (h) $h(0) = 1$ (i) $\lim\limits_{x \to 2} h(x) = 2$ (j) $h(2)$ does not exist (k) $\lim\limits_{x \to 5^{+}} h(x) = 3$ (l) $\lim\limits_{x \to 5^{-}} h(x)$ does not exist

Work Step by Step

(a) $h(x)$ approaches $y = 4$ as the function approaches $x = -3$ from the left. (b) $h(x)$ approaches $y = 4$ as the function approaches $x = -3$ from the right. (c) As $h(x)$ approaches $x = -3$ from both directions, it also approaches $y = 4$. (d) The function $h(x)$ has no $y$ value at $x = -3$, which is why there is a hole in the function at that point. (e) $h(x)$ approaches $y = 1$ as the function approaches $x = 0$ from the left. (f) $h(x)$ approaches $y = -1$ as the function approaches $x = 0$ from the right. (g) The limits from the left and from the right do not match. (h) Although $h(x)$ approaches $y = -1$ from the right, the function is only present at $y = 1$. (i) $h(x)$ approaches $y = 2$ as it approaches $x = 2$ from both sides. (j) $h(x)$ has no $y$ value at $x = 2$. (k) $h(x)$ approaches $y = 3$ as the function approaches $x = 5$ from the right. (l) $h(x)$ oscillates between $y = 2$ and $y = 4$, meaning that it is approaching no single value as it gets closer to $x = 5$ from the left.
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