{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Assignment3(1804)(Solution)(9-3-11)

# Assignment3(1804)(Solution)(9-3-11) - MATH 1804 University...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 11

Assignment3(1804)(Solution)(9-3-11) - MATH 1804 University...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online