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# HW02 - 2 β x 2-β x Β 1 β x 1 β x = 2 β x 1 2 β...

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MATH 132 HW 02 Answer Section 2.4 [18]. Find the limits (a) lim x 1 + 2 x ( x - 1) | x - 1 | = lim x 1 + 2 x ( x - 1) x - 1 = lim x 1 + 2 x = 2 x - 1 > 0 = | x - 1 | = x - 1 (b) lim x 1 - 2 x ( x - 1) | x - 1 | = lim x 1 - 2 x ( x - 1) - ( x - 1) = lim x 1 - - 2 x = - 2 x - 1 < 0 = | x - 1 | = - ( x - 1) [26]. lim t 0 2 t tan t = lim t 0 2 t sin t · cos t = lim t 0 2 t sin t · cos t = 2 · 1 · 1 = 2 tan t = sin t cos t [32]. lim h 0 sin(sin h ) sin h = 1 [46]. lim r →∞ r +sin r 2 r +7 - 5 sin r = 0 . 5 Solution: r + sin r 2 r + 7 - 5 sin r · 1 r 1 r = 1 + sin r r 2 + 7 r - 5 sin r r 1 + 0 2 + 0 - 0 = 1 2 as x → ∞ [58]. lim x →∞ 2+ x 2 - x = - 1 Solution:
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Unformatted text preview: 2 + β x 2-β x Β· 1 β x 1 β x = 2 β x + 1 2 β x-1 β 0 + 1-1 =-1 as x β β Note: 1. lim x β sin x x = 1 2. lim x β x sin x = lim x β 1 sin x x = lim x β 1 lim x β sin x x = 1 3. If lim x β c f ( x ) = 0, then lim x β c sin f ( x ) f ( x ) = 1...
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