By Moh T.T.

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**Additional info for Algebra**

**Example text**

Thus, either or is 0 for any value of . Then lim () and lim () do not exist, but lim [ ()()] = lim 0 = 0. →0 →0 →0 →0 63. Since the denominator approaches 0 as → −2, the limit will exist only if the numerator also approaches 0 as → −2. In order for this to happen, we need lim →−2 2 3 + + + 3 = 0 ⇔ 3(−2)2 + (−2) + + 3 = 0 ⇔ 12 − 2 + + 3 = 0 ⇔ = 15. With = 15, the limit becomes lim →−2 32 + 15 + 18 3( + 2)( + 3) 3( + 3) 3(−2 + 3) 3 = lim = lim = = = −1.

4−3 1 (b) Using the points (2 4) and (5 23) from the approximate tangent line, the instantaneous velocity at = 3 is about 23 − 4 ≈ 63 ms. 5−2 9. (a) For the curve = sin(10) and the point (1 0): 2 (2 0) 0 05 (05 0) 15 (15 08660) 17321 06 (06 08660) 14 (14 −04339) −10847 07 (07 07818) 08 (08 1) 43301 09 (09 −03420) 13 12 11 (13 −08230) (12 08660) (11 −02817) −27433 −28173 0 −21651 −26061 −5 34202 As approaches 1, the slopes do not appear to be approaching any particular value.

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). 13. True. Suppose that lim [ () + ()] exists.