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Unformatted text preview: MATH 1804 University Mathematics A Assignment 3 – Solution March 9, 2011 1. (a) lim x →∞ x 2 + 3 x 2 x 3 x + 1 = lim x →∞ 2 x + 3 6 x 2 1 (L’Hˆ opital’s Rule , ∞ ∞ ) = lim x →∞ 2 12 x (L’Hˆ opital’s Rule , ∞ ∞ ) = 0 (b) lim x → 1 (3 x + 1) √ x 4 x 2 3 1 = lim x → 1 3 x 3 2 + x 1 2 4 x 2 3 1 = lim x → 1 9 2 x 1 2 + 1 2 x 1 2 2 3 x 1 3 (L’Hˆ opital’s Rule , ) = 15 2 (c) lim x → sin x cos x p ( x ) = lim x → cos 2 x sin 2 x p ( x ) (L’Hˆ opital’s Rule , ) = 1 2 (d) lim x → π 2 (sec x tan x ) = lim x → π 2 1 sin x cos x = lim x → π 2 cos x sin x (L’Hˆ opital’s Rule , ) = 0 Remark on 1. (c) (For the rigour of the solution, it remains to) justify the following: (i) The use of the L’Hˆ opital’s Rule: How are the assumptions in Theorem 4.7 or Theorem 4.10 in the lecture notes satisfied (so that applying the Rule is valid)? (ii) In the answer, 1 in the numerator obviously comes from cos 2 sin 2 0 = 1 2 2 = 0. But why is it valid to take p (0) = 2 as the denominator? 1 (d) This limit can be evaluated without the L’Hˆ opital’s Rule: Note, by the idenity sin 2 x +cos 2 x = 1, that sec x tan x = 1 sin x cos x = 1 sin 2 x cos x (1 + sin x ) = cos x 1 + sin x , thus lim x → π 2 (sec x tan x ) = 1 + 1 = 0 . 2. (a) The vertical asymptotes are x = ± 3 , since lim x → 3 ± f ( x ) = ∓∞ and lim x → 3 ± f ( x ) = ±∞ . The other asymptote is y = 1 , since lim x →±∞ f ( x ) x = lim x →±∞ 10 x 2 9 x x 3 = lim x →±∞ 2 x 9 3 x 2 = lim x →±∞ 2 6 x = 0 , lim x →±∞ ( f ( x ) x ) = lim x →±∞ 10 x 2 9 x 2 = lim x →±∞ 2 x 2 x = 1 . (b) The vertical asymptote is x = 5 3 , since lim x → 5 3 ± f ( x ) = ±∞ . The other asymptote is y = 2 3 , since lim x →±∞ f ( x ) x = lim x →±∞ 2 x + 1 3 x 2 5 x = lim x →±∞ 2 6 x 5 = 0 , lim x →±∞ ( f ( x ) x ) = lim x →±∞ 2 x + 1 3 x 5 = lim x →±∞ 2 3 = 2 3 . (c) The vertical asymptote is x = 1 2 , since lim x → 1 2 ± f ( x ) = ∓∞ . The other asymptote is y = 5 2 x 5 4 , since lim x →±∞ f ( x ) x = lim x →±∞ 5 x 2 2 2 x 2 + x = lim x →±∞ 10 x 4 x + 1 = lim x →±∞ 10 4 = 5 2 , lim x →±∞ f ( x ) 5 2 x = lim x →±∞ 5 x 4 2(2 x + 1) = lim x →±∞ 5 4 = 5 4 . 2 3. f ( x ) = 7 x 2 36 x + 32 ( x + 1) 4 = (7 x 8)( x 4) ( x + 1) 4 > 0 on (∞ , 8 7 ) ∪ (4 , ∞ ) \ { 1 } , = 0 at 8 7 , 4 , < 0 on ( 8 7 , 4) . The critical points are 1 (if f is defined there), 8 7 and 4. From the table x (∞ , 1) ( 1 , 8 7 ) ( 8 7 , 4) (4 , ∞ ) f + + + f increasing increasing decreasing increasing we see that 8 7 is a local maximum point, and 4 is a local minimum point....
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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
 Math

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