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solution AS2 - 1-f x = 3 vii lim x → 1 f x = 3 viii lim x...

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Page 1 Math151 Assignment 2 - Section 2.2
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Page 1 Math151 Assignment 2 - Section 2.3
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Page 1 Math151 Assignment 2 - Section 2.4
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Assignment 2 Solutions Instructor’s Questions: 1. Let f ( x ) = 2 x | x | if x < - 2 - 1 if x = - 2 4 - x 2 if - 2 < x < 1 2 if x = 1 2 x + 1 if x > 1 (a) Sketch a graph of f . (b) Find f ( - 3), f ( - 2), f (0), f (1) and f (5). (c) Find the following limits, if they exist: i. lim x →-∞ f ( x ) ii. lim x →- 3 f ( x ) iii. lim x →- 2 - f ( x ) iv. lim x →- 2 + f ( x ) v. lim x →- 2 f ( x ) vi. lim x 1 - f ( x ) vii. lim x 1 + f ( x ) viii. lim x 1 f ( x ) ix. lim x →∞ f ( x ) 1
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Solution: Figure 1: Problem 1 (a) (b) f ( - 3) = - 2, f ( - 2) = - 1, f (0) = 4, f (1) = 2, and f (5) = 9. (c) Find the following limits, if they exist: i. lim x →-∞ f ( x ) = - 2 ii. lim x →- 3 f ( x ) = - 2 iii. lim x →- 2 - f ( x ) = - 2 iv. lim x →- 2 + f ( x ) = 0 v. lim x →- 2 f ( x ) Does not exist since the right-hand and the left- hand limits at x = - 2 are not equal. vi. lim x
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Unformatted text preview: 1-f ( x ) = 3 vii. lim x → 1 + f ( x ) = 3 viii. lim x → 1 f ( x ) = 3 ix. lim x →∞ f ( x ) = ∞ (This is more precise than ”The limit does not exist”, even the statement is true.) 2 2. Find constants a and b such that lim x → √ ax + b-2 x = 1 . Solution. Since the denominator is approaching 0, for this limit to exist the numerator must also approach 0. Thus 0 = lim x → ( √ ax + b-2) = √ b-2 ⇒ √ b = 2 ⇒ b = 4 . Next 1 = lim x → √ ax + 4-2 x = lim x → √ ax + 4-2 x · √ ax + 4 + 2 √ ax + 4 + 2 = lim x → ax x ( √ ax + 4 + 2) = lim x → a √ ax + 4 + 2 = a 4 Thus a = 4. 3...
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solution AS2 - 1-f x = 3 vii lim x → 1 f x = 3 viii lim x...

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