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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.

**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}.