Section 1.5 Limits at Infinity, Infinite Limits 81 SOLUTION We often have a vertical asymptote at a point where the denomi nator is zero, and in this case we do because 2x urn 2x . and 00 lim —----— = —CX) x—1 x—’i On the other hand, urn x—’ f(x) Fjire 2r and soy Figure 8. = = 2x x — 1 = 2 urn x— 1 =2 1/x — 2x urn and 2 -- _,—_cxJx 2 is a horizontal asymptote. The graph of y I — 2x/(x 1) is shown in — 8 (;ollcepts Review ii I. [o say that x / (x) = L means that C- means that to say that Give your answers in informal 3. If Iimf(x) I iiiiae. 4. If lim f(x) 2. To say that lim f(x) liii 1(x) ii = = means that — to say that means that 6, then the line = tote of the graph of y = tote of the graph of y = asymp is a asymp then the line 00, = is a f(x). f(x). Give your answers in infor language. Problem Set 1.5 In l’rohlems 1—42, find (lie limits. I. jim = x — 7 x(x - 2. in 5 4. (2 — 21. 5)(3 - x) 6. divide by \/2x 9. Iim ---= - 7TX 8. )2 2 5x urn jim 15 1+ 8x 2 x + 2 4 20 sin 10. lim 5—*(XJ 02 5 16. 18. n’ 1 7 In Iimj In 5. — — . cxy2y+2 I-lint: Divide numerator and denomi — + X 0 (1 x” a 1 + - + a _ x + a 1 + + b_ 1 x + b, a - lim— 26. i where a 0 — + 28. lim (—‘--_; 29. lim + 30. — 9 2n + I 9 + 3 12 X lim lim “== l)(x + I) lim , “ x” + b 0 h x” 1 + 27. (x lim x—’c 0, and n is a natural number. 25. Hint: Divide numerator and denominator \/‘x + 3 by x. Note that, for x > 0, \/ + 3/i = \/(x2 + 3)/x . 2 19. 3 + I 9y lim + 7x 2 a ,2 24. — -s lim .r—’V’ 5 — (32 31. 33. - + \/2x . urn v-’a (x jim x-- 3 x )(3 — — — 3 x) 32. 34. lim sin C) irO Iirn v—’( r/2)’ COS C) v-Xr 20. lim x+4 and nator by y . 2 b() — 14. 3 Mulliply 1-lint: 1im(\/+2x_•r) Y’ SO — 12. 15. 11m 2 17. ii + + \/_5). — T0 3\/3 limq 23. — — II. Jim 13. 22. Hrn— - 2xIOOx lim (V cx) 35. 37. limXX x—3 36. jim urn 38. x—’3 r—’O X • jim r +2x—8 -——-,--—-——---- x—4 Ix 0,