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Pre-Calc Homework Solutions 77

# Pre-Calc Homework Solutions 77 - Section 3.3 1 x 77 31(a...

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31. (a) Note that 2 x # sin } 1 x } # x , for all x , so lim x 0 1 x sin } 1 x } 2 5 0 by the Sandwich Theorem. Therefore, f is continuous at x 5 0. (b) } f (0 1 h h ) 2 f (0) } 5 5 sin } 1 h } (c) The limit does not exist because sin } 1 h } oscillates between 2 1 and 1 an infinite number of times arbitrarily close to h 5 0 (that is, for h in any open interval containing 0). (d) No, because the limit in part (c) does not exist. (e) } g (0 1 h h ) 2 g (0) } 5 5 h sin } 1 h } As noted in part (a), the limit of this as x approaches zero is 0, so g 9 (0) 5 0. Section 3.3 Rules for Differentiation (pp. 112–121) Quick Review 3.3 1. ( x 2 2 2)( x 2 1 1 1) 5 x 2 x 2 1 1 x 2 ? 1 2 2 x 2 1 2 2 ? 1 5 x 1 x 2 2 2 x 2 1 2 2 2. 1 } x 2 1 x 1 } 2 2 1 5 } x 2 1 x 1 } 5 } x x 2 } 1 } 1 x } 5 x 1 x 2 1 3. 3 x 2 2 } 2 x } 1 } x 5 2 } 5 3 x 2 2 2 x 2 1 1 5 x 2 2 4. } 3 x 4 2 2 2 x x 2 3 1 4 } 5 } 3 2 x x 4 2 } 2 } 2 2 x x 3 2 } 1 } 2 4 x 2 } 5 } 3 2 } x 2 2 x 1 2 x 2 2 5. ( x 2 1 1 2)( x 2 2 1 1) 5 x 2 1 x 2 2 1 x 2 1 ? 1 1 2 x 2 2 1 2 ? 1 5 x 2 3 1 x 2 1 1 2 x 2 2 1 2 6.
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