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**Extra info for Instructor's Solutions Manual to Calculus: One and Several Variables**

**Sample text**

Then c − δ < c − |h| < c and, by (∗), |f (c − |h|) − L | < . Thus lim f (c − |h|) = L. h→0 Conversely we now assume that lim f (c − |h|) = L. Then for h→0 (∗∗) if 0 < |h| < δ then > 0 there exists δ > 0 such that |f (c − |h|) − L | < . Suppose now that c − δ < x < c. Then 0 < c − x < δ so that, by (∗∗), |f (c − (c − x)) − L | = |f (x) − L | < . Thus lim f (x) = L. x→c− 57. We begin by assuming that lim f (x) = L and showing that + x→c lim f (c + |h|) = L. h→0 Let > 0. Since lim f (x) = L, there exists δ > 0 such that + x→c (∗) if c < x < c + δ then |f (x) − L | < .

Lim x + x→1 1 x =2 29. does not exist 31. 2x − 5x2 = lim (2 − 5x) = 2 x→0 x→0 x lim 33 15:53 P1: PBU/OVY JWDD027-02 P2: PBU/OVY 34 32. 1 lim x→3 x−3 x−3 1 = lim = lim − 6 − 2x x→3 2(3 − x) x→3 2 =− 1 2 33. x2 − 1 (x − 1)(x + 1) = lim = lim (x + 1) = 2 x→1 x − 1 x→1 x→1 x−1 34. x3 − 1 (x − 1)(x2 + x + 1) = lim = lim x2 + x + 1 = 3 x→1 x − 1 x→1 x→1 (x − 1) 35. 0 36. does not exist 37. 1 38. 3 39. 16 40. 0 41. does not exist 42. 2 43. 4 44. 0 45. does not exist 46. 2 lim lim 47. √ lim x→1 x2 + 1 − x−1 √ 2 √ √ √ √ ( x2 + 1 − 2 )( x2 + 1 + 2 ) √ √ x→1 (x − 1)( x2 + 1 + 2 ) = lim x2 − 1 x+1 1 2 √ √ = lim √ √ = √ =√ x→1 (x − 1)( x2 + 1 + 2 2 2 2 ) x→1 x2 + 1 + 2 = lim 48.

17. 18. 19. 20. 22. 23. 24. 21. Impossible Impossible Impossible 25. If f (0) = 0 or if f (1) = 1, we have a ﬁxed point. If f (0) = 0 and f (1) = 1, then set g(x) = x − f (x). g is continuous on [ 0, 1 ] and g(0) < 0 < g(1). Therefore there exists a number c ∈ (0, 1) such that g(c) = c − f (c) = 0. 26. Set h(x) = f (x) − g(x). Then h is continuous on [a, b], and h(a) = f (a) − g(a) < 0, h(b) = f (b) − g(b) > 0. By the intermediate value theorem there exists a number c ∈ (a, b) such that h(c) = 0.