By Loyer, Triola

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**Example text**

2 (iii) [[]] = −3 for −3 ≤ −2, so lim [[]] = →−24 lim (−3) = −3. →−24 (b) (i) [[]] = − 1 for − 1 ≤ , so lim [[]] = lim ( − 1) = − 1. →− →− (ii) [[]] = for ≤ + 1, so lim [[]] = lim = . →+ →+ 33 34 ¤ CHAPTER 1 FUNCTIONS AND LIMITS (c) lim [[]] exists ⇔ is not an integer. → 53. The graph of () = [[]] + [[−]] is the same as the graph of () = −1 with holes at each integer, since () = 0 for any integer . Thus, lim () = −1 and lim () = −1, so lim () = −1.

Sin(2): Start with the graph of = sin and stretch horizontally by a factor of 2. 19 20 ¤ 17. = CHAPTER 1 FUNCTIONS AND LIMITS 1 2 (1 − cos ): Start with the graph of = cos , reflect about the -axis, shift 1 unit upward, and then shrink vertically by a factor of 2. 19. = 1 − 2 − 2 = −(2 + 2) + 1 = −(2 + 2 + 1) + 2 = −( + 1)2 + 2: Start with the graph of = 2 , reflect about the -axis, shift 1 unit to the left, and then shift 2 units upward. 21. = | − 2|: Start with the graph of = || and shift 2 units to the right.

Then ( + ) = (−1 + 1)2 = 02 = 0, but () + () = (−1)2 + 12 = 2 6= 0 = ( + ). 3. False. Let () = 2 . Then (3) = (3)2 = 92 and 3 () = 32 . So (3) 6= 3 (). 46 ¤ CHAPTER 1 FUNCTIONS AND LIMITS 5. True. See the Vertical Line Test. 7. False. Limit Law 2 applies only if the individual limits exist (these don’t). 9. True. Limit Law 5 applies. 11. False. Consider lim →5 ( − 5) sin( − 5) or lim . The first limit exists and is equal to 5. 5, →5 −5 −5 we know that the latter limit exists (and it is equal to 1).