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1_ y = 4, then lim(x 2 -1- x = 5x] = 1. lim(2x + 1) 2. g(x)] = iimf(x) — iiing(x) iirnf(x) -I- (—1)liing(x) urn f(x) -I- iim(—1)g(x) urn [f(x) -+ (—1)g(x)] Squeeze Theorem 2 such that 0 < Is — ,6 1 inin{6 , 2 Is cI <6 L. L n 63 — < f(x) <f(x) : g(x) g(x) ‘ = ......_.. x—0 (x) 2 f —2, then lirn— x—c g(r) = = 1 = lim (3x 2 — 1) — X 4. If lirnf(x) x-’O = x—2 5 — 3x 2x+1 lim [(2x 2+ 5 1mm 4. 1)(7x2 -I- 3. lirn [(2x ± 1)(x — 3)] x—0 I 6. 13)] Land tim g(x) = = sinx 1mm— iim[f(x)-Lg(x) x—O = X 1 for C, 1. It follows —-3 7 — 2,2 4x-i- 1 L, then lim = I /6 g(s) = (sin x)/x, and h(s) 2 x him h(s) and so, by Theorem B, SOLUTION Let f(x) that lini f(x) r-O (sin x)/x h(x) < L + e h(x) Assume that we have proved 1 — 5216 = - cI all x near but different from 0. What can we conclude about urn • EXAMPLE 9 — 63}. Then 0< L—e<h(x)<L-+c 2 0<lx—cI<6 We conclude that lim g(x) Let6 Choose 63 so that and 0<lx—cI<i=L—s<f(x)<L-+c Proof (Optional) Let c > 0 be given. Choose 6 such that 3)f(x) = — g(r) h(.x) for all x near Let f, g, and h be functions satisfying f(.x) except possibly at c.If iirnf(x) = liinh(x) = L, then linlg(.x) = L. Theorem B In Problems 1—12, use Theorem A to find each of the limits. Justify each step by appealing to a numbered statement, as in Examples 1—4. + x—’c 4 and tim g(x) Problem Set 1.3 x—c 3. If limf(x) [f(s) The Squeeze Theorem You have likely heard someone say, “I was caught between a rock and a hard place.” This is what happens to g in the following theorem (see Figure 2), lim Proof of Statement 5 g(x)=_2,thenh 2 lfih V g2(x) + 12= 2. 9 1. If limf(x) Concepts Review C Chapter 1 Limits Figure 2 L 72 — 3 ± 9w 19)_h/2 10. 8. -i-2x 2 V5x urn \/—3w -)- 7w 2 urn urn w--2 -)--4w+4 2 w iiin[If(t)I -I- 13g(t)I] = = 5 1 — 32 34. f(s) = = = = jg(x)IIf(x) — LI — < Is — cI <6 = Ig(x)I < Ml + 1 M, then there is a number 5 such = 0 L cl. = = [f(x) = 0. = 0. g(x) exists; X 4 + 4x + x 3 48. urn [x 2 + 2x] 46. hm (x — [x]]) x-i- urn x__ir+ 44. urn 42. = 0. perimeter of R perimeter of Q Answers to Concepts Review: 1. 48 2. 4 3. —8;—4+5c 4.0 51. Let y = V and consider the points M, N, 0, and P with coordinates (1,0), (0,1), (0,0), and (x, y) on the graph of y = V, respectively. Calculate perimeter of tSNOP area of ANOP (a) jim jim x—o perimeter of t\MOP (b) x—O area of 1M0P lirn x—O 50. Let R be the rectangle joining the midpoints of the sides of the quadrilateral Q having vertices (±x, 0) and (0, ±1). Calculate 49. Suppose that f(x)g(x) = 1 for all x and urn g(x) x a Prove that lim f(x) does not exist. IxI (3x — 1) 2 (x2+1)x] ‘—3 47• jim -x—O x—2 }+ hm x_*_3* 45 jim 43. 41. In Problems 41—48,find each of the right-hand and left-hand limits or state that they do not exist. iimf(x) or urn g(x) exists. (b) jim [f(x) g(x)] exists, this does not imply that either . limlf(x)I iim[f(x) — LI + g(x)] exists, this does not imply that either iimf(x) or urn (a) urn 40. Find examples to show that if 39. Provethatlimlxi 38. Provethatiimf(x) 37. Provethatiirnf(x) 36. Prove Statement 7 of Theorem A by first giving an c—b proof that urn [1/g(x)] = i/[lim g(x)] and then applying State ment 6. 0 Now show that if urn g(x) that 73 Theorem B of the previous section says that limits of polynomial functions can always be found by substitution, and limits of rational functions can be found by substitution as long as the denominator is not zero at the limit point. This substitu tion rule applies to the trigonometric functions as well. This result is stated next. -f- Mu ILIIg(x) MI — — Lg(x) -f- Lg(x) — LMI Ig(x)[f(x) — L] + L[g(x) If(x)g(x) 1.4 Limits Involving — LMI Trigonometric Functions If(x)g(x) x — 3 2 -I- 2x + 1 3x f(2)1/(x — 2) for each 32. f(s) — 35. Prove Statement 6 of Theorem A. ifint: 33. f(x) 31. f(x) 3]4 30. lim[f(u) + 3g(u)j 3 28. hrn[f(i) In Problems 31—34, find urn [f(x) given functionf 29. 27. IiinY[f(x) + — in Problems 25—3U, find the limits if Jim f(s) = 3 and urn g(x) = —1 (see Example 4). 2f(x)—3g(x) 25. limn\/f2(x) -I- g (x) 2 26. lien x—a f(x) -I-- g(x) 24. — (w + 2)(w 2 — w — 6) — + 1 + — -1- 6 -x 2 1-x—2 — 14x 51 20. lini x—’—3 x —1 2 x 2 — 4x — 21 ux±2u2x 2 u +uxxu 2 ,x 22. urn 21. urn u——2 x—1 2 u u 6 2 + 2x — 3 x -I2 — 6x1r 4’ir 2x 23. Jim x—’l 19. Urn — 2 5x x—2 — x—-1 2 16. jim x—2 14. urn 6x + lix — 6 2 17. jim x——1 x 3 + 4x 2 — 19i —I- 14 2 + 7x + 10 x 18. liin 2 xx+2 x—1 15. jim — 2x —3 x+1 2 X x— 2 -I- 4 13. Jim in Problems 13—24, find the indicated limit or state that it does not exist. In many cases, you will want to do some algebra before trying to evaluate the limit. 5 w-. 4 12. lim(2w ) 15)13 Iim”’_i 11. y2 y±4 3 -I9. urn (2t 7. ijrnV’ Section 1.4 Limits Involving Trigonometric Functions