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Unformatted text preview: Suggested Solutions to Assignment 2 MATH 1804 UNIVERSITY MATHEMATICS A 1. Solution . Compute the following limits: (a) lim x → 3 x + 3 √ x 2 6 √ 3 = lim x → 3 x + 3 √ x 2 6 √ 3 √ x 2 6 + √ 3 √ x 2 6 + √ 3 = lim x → 3 ( x + 3)( √ x 2 6 + √ 3) x 2 9 = lim x → 3 √ x 2 6 + √ 3 x 3 = √ 3 3 Another method: By L H ˆ opital s rule, we have lim x → 3 x + 3 √ x 2 6 √ 3 = lim x → 3 1 ( x 2 6) 1 2 · x = √ 3 3 (b) lim x → 1 + 1 x 1 + 1  x 1  = lim x → 1 + 1 x 1 + 1 x 1 = lim x → 1 + 2 x 1 = ∞ (c) lim x → 1 1 x 1 + 1  x 1  = lim x → 1 1 x 1 1 x 1 = lim x → 1 + 0 = 0 2. Solution . Use the Sandwich Theorem to compute the following limits: (a) lim x → x 4 cos( 4 π x ) Notice that the cosine function satisfies that 1 ≤ cos y ≤ 1 for any y ∈ R . As long as x 6 = 0, we also have 1 ≤ cos 4 π x ≤ 1 1 Then multiply the above inequality by x 4 and get x 4 ≤ x 4 cos 4 π x ≤ x 4 Now, since lim x → x 4 = lim x → x 4 = 0 It follows from the Sandwich Theorem that lim x → x 4 cos 4 π x = 0 (b) lim x →∞ sin( x 2 ) x 2 Notice that the sine function satisfies that 1 ≤ sin( x 2 ) ≤ 1 for any x ∈ R . Multiply the above inequality by 1 x 2 and we get 1 x 2 ≤ 1 x 2 sin( x 2 ) ≤ 1 x 2 Since lim x →∞ 1 x 2 = lim x →∞ 1 x 2 = 0 It follows from the Sandwich Theorem that lim x...
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This note was uploaded on 04/19/2011 for the course MATH 1804 taught by Professor Ng during the Spring '11 term at HKU.
 Spring '11
 Ng
 Limits

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