Download Calculus: Early Transcendentals (6th) -- Student Solutions by James Stewart PDF

By James Stewart

Offers thoroughly worked-out strategies to all odd-numbered workouts in the textual content, giving scholars how to fee their solutions and confirm that they took the right kind steps to reach at a solution.

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25. Given ε > 0, we need δ > 0 such that if 0 < |x − 0| < δ, then x2 − 0 < ε Then 0 < |x − 0| < δ ⇒ ⇔ x2 < ε ⇔ |x| < √ √ ε. Take δ = ε. x2 − 0 < ε. Thus, lim x2 = 0 by the definition of a limit. x→0 27. Given ε > 0, we need δ > 0 such that if 0 < |x − 0| < δ, then |x| − 0 < ε. But |x| = |x|. So this is true if we pick δ = ε. Thus, lim |x| = 0 by the definition of a limit. x→0 54 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 29. Given ε > 0, we need δ > 0 such that if 0 < |x − 2| < δ, then (x − 2)2 < ε. So take δ = √ ε.

H→0 h→0 (⇐) Let ε > 0. Since lim f(a + h) = f (a), there exists δ > 0 such that 0 < |h| < δ h→0 ⇒ |f (a + h) − f (a)| < ε. So if 0 < |x − a| < δ, then |f (x) − f (a)| = |f (a + (x − a)) − f (a)| < ε. Thus, lim f (x) = f (a) and so f is continuous at a. x→a 57. As in the previous exercise, we must show that lim cos(a + h) = cos a to prove that the cosine function is continuous. h→0 lim cos(a + h) = lim (cos a cos h − sin a sin h) = lim (cos a cos h) − lim (sin a sin h) h→0 h→0 = 59. f (x) = lim cos a h→0 0 if x is rational 1 if x is irrational h→0 lim cos h − lim sin a h→0 h→0 h→0 lim sin h = (cos a)(1) − (sin a)(0) = cos a h→0 is continuous nowhere.

Thus, we + π2 n. 90). Just as 150◦ is the reference angle for 30◦ , π − sin−1 reference angle for sin−1 π4 . 24). 3 Calculating Limits Using the Limit Laws 1. 3 (d) lim x→2 lim [3f (x)] 3f (x) x→2 = g(x) lim g(x) CALCULATING LIMITS USING THE LIMIT LAWS ¤ 47 [Limit Law 5] x→2 3 lim f (x) = x→2 [Limit Law 3] lim g(x) x→2 = 3(4) = −6 −2 (e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim x→2 g(x) , does not exist because the h(x) denominator approaches 0 while the numerator approaches a nonzero number.

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