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

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

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

**Sample text**

0 1. 1 2 3. x(1 + x) 1+x ; = lim 2 x→0 x→0 2x 2x 4. 15. lim lim f (x) = lim (x2 − x) = 2 1 18. 21. δ1 22. 25. 02 26. x(1 + x) = 0 9 2 lim does not exist 16. x→2+ 17. 5 x→1 19. 1 23. δ= 27. 75 29. 1 30. 03 31. 43 32. 028 33. 14 34. 05 20. 10 24. δ= 28. 05 35. Since |(2x − 5) − 3| = |2x − 8| = 2|x − 4|, we can take δ = 1 2 : if 0 < |x − 4| < 1 2 then, |(2x − 5) − 3| = 2|x − 4| < . 2 36. Since |(3x − 1) − 5| = |3x − 6| = 3|x − 2|, we can take δ = 1 3 : if 0 < |x − 2| < 1 3 then, |(3x − 1) − 5| = 3|x − 2| < .

A) If f (x) ≥ g(x) then |f (x) − g(x)| = f (x) − g(x) and 1 2 {[f (x) + g(x)] + |f (x) − g(x)|} = 1 2 {f (x) + g(x) + f (x) − g(x)} = 1 2 · 2f (x) = f (x) = max {f (x), g(x)}. If f (x) ≤ g(x) then |f (x) − g(x)| = −[f (x) − g(x)] = g(x) − f (x) and 1 2 {[f (x) + g(x)] + |f (x) − g(x)|} = 1 2 {f (x) + g(x) + g(x) − f (x)} = 1 2 · 2g(x) = g(x) = max {f (x), g(x)}. (b) min {f (x), g(x)} = 53. If lim f (x) = L x→c 1 2 {[f (x) + g(x)] − |f (x) − g(x)|} and lim g(x) = L, 1 {[f (x) x→c 2 lim h(x) = lim x→c then x→c 1 [f (x) x→c 2 = lim + g(x)] − |f (x) − g(x)|} | (x) x→c f + g(x)] − lim − g(x)| = 12 (L + L) − 12 (L − L) = L.

Then there exists δ > 0 such that |f (x) − L| < then Now suppose that lim [f (x) − L] = 0 and let 0 < |x − c| < δ, |f (x) − L − 0| < then lim [f (x) − L] = 0. x→c > 0. Then there exists δ > 0 such that x→c if =⇒ =⇒ lim |f (x) − L| < x→c which implies lim f (x) = L x→c = L. Since lim f (x) = L, there exists δ > 0 such that if 0 < |x − c| < δ then 59. (a) Let x→c L − f (x) ≤ |L − f (x)| = |f (x) − L| < L Therefore, f (x) > L − L = 0 for all x ∈ (c − δ, c + δ); take γ = δ. = −L and repeat the argument in part (a).