By Saturnino L. Salas, Garret J. Etgen, Einar Hille

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**Read or Download Instructor's Solutions Manual to Calculus: One and Several Variables PDF**

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**Additional resources for Instructor's Solutions Manual to Calculus: One and Several Variables**

**Sample text**

Jump discontinuity 12. removable discontinuity 13. continuous 14. indeﬁnite discontinuity 15. inﬁnite discontinuity 16. 17. inﬁnite discontinuity 18. y y 3 2 1 1 -1 x 1 -1 no discontinuities x 1 no discontinuities 19. 20. y y 10 4 2 -3 -2 -2 -1 1 2 3 x 2 -10 inﬁnite discontinuity at −3 no discontinuities 21. 22. 4 23. 24. jump discontinuity at −2 no discontinuities 25. 26. jump discontinuity at −1 jump discontinuities at 0 and 2 27. 28. 4 29. 30. (One possibility) (One possibility) 31. f (1) = 2 32.

X→2 Now, =⇒ x→1+ lim f (x) = lim f (x) + x→2− iﬀ x→2+ that is, iﬀ 4B − A = 6. More simply we have A − B = 3 are unequal; 47. x = 0, lim f (x) = 3 x→1+ 4B − A = 6 and B = 3. 41. f (5) = 1 6 44. f (5) = 1 4 continuous only at x = 0. 4 49 48. (a) (b) Hc is continuous on (−∞, c) ∪ [c, ∞). P ,c is continuous on (−∞, c) ∪ [c, c + ) ∪ [c + , ∞). δc is continuous on (−∞, c) ∪ (c, ∞). 49. 6). Use the equivalence of (i) and (ii) setting L = f (c). 50. (a) Let = f (c) > 0. By the continuity of f at c, there exists δ > 0 such that |f (x) − f (c)| < f (c) for all x ∈ (c − δ, c + δ).

A) Set δ = √ c. By the hint, if (b) Set δ = 2 0 < |x − c| < . If 0 < x < 2 , √ then 50. Take δ = minimum of 1 and /5. √ √ 1 then | x − c | < √ |x − c| < . c √ √ | x − 0| = x < . c If 0 < |x − 2| < δ, then |x − 2| < /5, |x + 2| < 5 and x − 4 = |x − 2| |x + 2| < ( /5)(5) = . 2 Therefore |x3 − 1| = |x2 + x + 1| |x − 1| < 7|x − 1| < 7( /7) = . 51. Take δ = minimum of 1 and /7. If 0 < |x − 1| < δ, then 0