Download Student Solutions Manual for Stewart's Single Variable by Jeffery Cole, Daniel Drucker, Daniel Anderson PDF

By Jeffery Cole, Daniel Drucker, Daniel Anderson

Presents thoroughly worked-out suggestions to all odd-numbered routines in the textual content, giving scholars the way to money their solutions and make sure that they took the right kind steps to reach at a solution.

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Read or Download Student Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals (6th Edition) PDF

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Extra resources for Student Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals (6th Edition)

Sample text

4. 76. 44. 76. 44. 44, or any smaller positive number. 5. 1106. 0906. 0906 (or any smaller positive number) since this is the smaller of δ1 and δ 2 . 52 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 7. For ε = 1, the definition of a limit requires that we find δ such that 4 + x − 3x3 − 2 < 1 1 < 4 + x − 3x3 < 3 ⇔ whenever 0 < |x − 1| < δ. 11. 11 (or any smaller positive number). 1 whenever 0 < |x − 1| < δ. 012. 012 (or any smaller positive number) for the inequality to hold. 9. 602 for x near π . 031 for M = 1000.

X→3 41. 5. 5− 2x − 1 2x − 1 −1 −1 −1 = lim = lim = −4. 25 50 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 1 1 − x |x| 43. Since |x| = −x for x < 0, we have lim x→0− = lim x→0− denominator approaches 0 and the numerator does not. 45. (a) 1 1 − x −x = lim x→0− 2 , which does not exist since the x (b) (i) Since sgn x = 1 for x > 0, lim sgn x = lim 1 = 1. x→0+ x→0+ (ii) Since sgn x = −1 for x < 0, lim sgn x = lim −1 = −1. x→0− x→0− (iii) Since lim sgn x 6= lim sgn x, lim sgn x does not exist. x→0− x→0 x→0+ (iv) Since |sgn x| = 1 for x 6= 0, lim |sgn x| = lim 1 = 1.

Given ε > 0, we need δ > 0 such that if 0 < |x − 2| < δ, then (x − 2)2 < ε. So take δ = √ ε. Then 0 < |x − 2| < δ x2 − 4x + 5 − 1 < ε ⇔ ⇔ |x − 2| < √ ε ⇔ x2 − 4x + 4 < ε ⇔ (x − 2)2 < ε. Thus, lim x2 − 4x + 5 = 1 by the definition of a limit. x→2 31. Given ε > 0, we need δ > 0 such that if 0 < |x − (−2)| < δ, then x2 − 1 − 3 < ε or upon simplifying we need x2 − 4 < ε whenever 0 < |x + 2| < δ. Notice that if |x + 2| < 1, then −1 < x + 2 < 1 ⇒ −5 < x − 2 < −3 ⇒ |x − 2| < 5. So take δ = min {ε/5, 1}.

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